Trigonometry

1. Trigonometry

ID is: 899 Seed is: 8584

Working with trigonometric equations: cosine

The diagram below shows a right-angled triangle. One angle is labelled $\text{57} \text{°}$ and one side is $\text{10}$ units long. Another one of the sides is labelled $x$ .

Solve for $x$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

x =

$x =$

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.

STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the adjacent side and the hypotenuse. So the trigonometric ratio to use is $\cos$ . We can write an equation with the given information and $x$ :

\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos 57 ° & = \frac{10}{x} \end{aligned}

$\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos 57° & = \frac{10}{x} \end{aligned}$

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for $x$ . Rearrange the equation to make $x$ the subject. Then evaluate the right side.

\begin{aligned} \cos 57 ° & = \frac{10}{x} \\ x & = \frac{10}{\cos 57 °} \\ = \frac{10}{0.54463...} \\ = 18.36078... \\ \approx 18.36 \end{aligned}

$\begin{aligned} \cos 57° & = \frac{10}{x} \\ x & = \frac{ 10 }{ \cos 57° } \\ & = \frac{ 10 }{ \text{0.54463...} } \\ & = \text{18.36078...} \\ & \approx \text{18.36} \end{aligned}$

The missing side, $x$ , is $\text{18.36}$ .

Submit your answer as:

ID is: 899 Seed is: 4000

Working with trigonometric equations: cosine

The diagram below shows a right-angled triangle. One angle is labelled $\text{49} \text{°}$ and one side is $\text{7}$ units long. Another one of the sides is labelled $x$ .

Solve for $x$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

x =

$x =$

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.

STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos 49 ° & = \frac{x}{7} \end{aligned}

$\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos 49° & = \frac{x}{7} \end{aligned}$

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for $x$ . Multiply by $\text{7}$ to isolate $x$ . Then evaluate the left side of the equation.

\begin{aligned} \cos 49 ° & = \frac{x}{7} \\ (7) \cos 49 ° & = x \\ (7) (0.65605...) & = x \\ 4.59241... & = x \\ 4.59 & \approx x \end{aligned}

$\begin{aligned} \cos 49° & = \frac{x}{7} \\ \left( 7 \right) \cos 49° & = x \\ \left( 7 \right) \left( \text{0.65605...} \right) & = x \\ \text{4.59241...} & = x \\ \text{4.59} & \approx x \end{aligned}$

The missing side, $x$ , is $\text{4.59}$ .

Submit your answer as:

ID is: 899 Seed is: 488

Working with trigonometric equations: cosine

The diagram below shows a right-angled triangle. One angle is labelled $\text{66} \text{°}$ and one side is $\text{9}$ units long. Another one of the sides is labelled $x$ .

Solve for $x$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

x =

$x =$

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.

STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos 66 ° & = \frac{9}{x} \end{aligned}

$\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos 66° & = \frac{9}{x} \end{aligned}$

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for $x$ . Rearrange the equation to make $x$ the subject. Then evaluate the right side.

\begin{aligned} \cos 66 ° & = \frac{9}{x} \\ x & = \frac{9}{\cos 66 °} \\ = \frac{9}{0.40673...} \\ = 22.12734... \\ \approx 22.13 \end{aligned}

$\begin{aligned} \cos 66° & = \frac{9}{x} \\ x & = \frac{ 9 }{ \cos 66° } \\ & = \frac{ 9 }{ \text{0.40673...} } \\ & = \text{22.12734...} \\ & \approx \text{22.13} \end{aligned}$

The missing side, $x$ , is $\text{22.13}$ .

Submit your answer as:

ID is: 922 Seed is: 6072

Trigonometry: using a calculator

Use your calculator to evaluate the following expression: $\; \sin54°$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

\sin 54 ° \approx

$\sin54° \approx$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to use your calculator to evaluate $\sin54°$ . If you are not sure about how to do that, refer to the manual of your calculator.

STEP: Use your calculator to evaluate

\sin 54 °

$\sin54°$ and then round off to two decimal places
[−2 points ⇒ 0 / 2 points left]

To determine the answer, we must use a calculator.

\begin{aligned} \sin 54 ° & = 0.80901... \\ \approx 0.81 \end{aligned}

$\begin{aligned} \sin54° & = \text{0.80901...} \\ & \approx \text{0.81} \end{aligned}$

The correct answer is $\text{0.81}$ .

Submit your answer as:

ID is: 922 Seed is: 9576

Trigonometry: using a calculator

Use your calculator to evaluate the following expression: $\; \sin74°$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

\sin 74 ° \approx

$\sin74° \approx$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to use your calculator to evaluate $\sin74°$ . If you are not sure about how to do that, refer to the manual of your calculator.

STEP: Use your calculator to evaluate

\sin 74 °

$\sin74°$ and then round off to two decimal places
[−2 points ⇒ 0 / 2 points left]

To determine the answer, we must use a calculator.

\begin{aligned} \sin 74 ° & = 0.96126... \\ \approx 0.96 \end{aligned}

$\begin{aligned} \sin74° & = \text{0.96126...} \\ & \approx \text{0.96} \end{aligned}$

The correct answer is $\text{0.96}$ .

Submit your answer as:

ID is: 922 Seed is: 5447

Trigonometry: using a calculator

Use your calculator to evaluate the following expression: $\; \sin74°$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

\sin 74 ° \approx

$\sin74° \approx$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to use your calculator to evaluate $\sin74°$ . If you are not sure about how to do that, refer to the manual of your calculator.

STEP: Use your calculator to evaluate

\sin 74 °

$\sin74°$ and then round off to two decimal places
[−2 points ⇒ 0 / 2 points left]

To determine the answer, we must use a calculator.

\begin{aligned} \sin 74 ° & = 0.96126... \\ \approx 0.96 \end{aligned}

$\begin{aligned} \sin74° & = \text{0.96126...} \\ & \approx \text{0.96} \end{aligned}$

The correct answer is $\text{0.96}$ .

Submit your answer as:

ID is: 898 Seed is: 8737

Working with trigonometric equations: tangent

The figure here shows a right-angled triangle. One angle is labelled $\text{43} \text{°}$ and one side is $\text{5}$ units long. Another one of the sides is labelled $x$ .

Calculate $x$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

x =

$x =$

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.

STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the opposite and adjacent sides. So the trigonometric ratio to use is $\tan$ . We can write an equation with the given information and $x$ :

\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \\ \tan 43 ° & = \frac{5}{x} \end{aligned}

$\begin{aligned} \tan \theta & = \frac{\text{opposite}}{\text{adjacent}}\\ \tan 43° & = \frac{5}{x} \end{aligned}$

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for $x$ . Start by rearranging the equation to make $x$ the subject. That means $\tan 43°$ will end up in the denominator on the right side.

\begin{aligned} \tan 43 ° & = \frac{5}{x} \\ x & = \frac{5}{\tan 43 °} \\ = \frac{5}{0.93251...} \\ = 5.36184... \\ \approx 5.36 \end{aligned}

$\begin{aligned} \tan 43° & = \frac{5} {x} \\ x & = \frac{ 5 }{ \tan 43° } \\ & = \frac{ 5 }{ \text{0.93251...} } \\ & = \text{5.36184...} \\ & \approx \text{5.36} \end{aligned}$

Therefore $x$ is $\text{5.36}$ .

Submit your answer as:

ID is: 898 Seed is: 1619

Working with trigonometric equations: tangent

The figure here shows a right-angled triangle. One angle is labelled $\text{51} \text{°}$ and one side is $\text{8}$ units long. Another one of the sides is labelled $x$ .

Determine the value of $x$ .

INSTRUCTION: Give your answer rounded to two decimal places.

Answer:

x =

$x =$

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.

STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \\ \tan 51 ° & = \frac{x}{8} \end{aligned}

$\begin{aligned} \tan \theta & = \frac{\text{opposite}}{\text{adjacent}}\\ \tan 51° & = \frac{x}{8} \end{aligned}$

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for $x$ . We need to multiply the $\text{8}$ across to the left side to isolate $x$ . Then evaluate the left side.

\begin{aligned} \tan 51 ° & = \frac{x}{8} \\ (8) \tan 51 ° & = x \\ (8) (1.23489...) & = x \\ 9.87917... & = x \\ 9.88 & \approx x \end{aligned}

$\begin{aligned} \tan 51° & = \frac{x} {8} \\ \left( 8 \right) \tan 51° & = x \\ \left( 8 \right) \left( \text{1.23489...} \right) & = x \\ \text{9.87917...} & = x \\ \text{9.88} & \approx x \end{aligned}$

Therefore $x$ is $\text{9.88}$ .

Submit your answer as:

ID is: 898 Seed is: 4028

Working with trigonometric equations: tangent

The figure here shows a right-angled triangle. One angle is labelled $\text{54} \text{°}$ and one side is $\text{7}$ units long. Another one of the sides is labelled $x$ .

Calculate $x$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

x =

$x =$

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.

STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \\ \tan 54 ° & = \frac{7}{x} \end{aligned}

$\begin{aligned} \tan \theta & = \frac{\text{opposite}}{\text{adjacent}}\\ \tan 54° & = \frac{7}{x} \end{aligned}$

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for $x$ . Start by rearranging the equation to make $x$ the subject. That means $\tan 54°$ will end up in the denominator on the right side.

\begin{aligned} \tan 54 ° & = \frac{7}{x} \\ x & = \frac{7}{\tan 54 °} \\ = \frac{7}{1.37638...} \\ = 5.08579... \\ \approx 5.09 \end{aligned}

$\begin{aligned} \tan 54° & = \frac{7} {x} \\ x & = \frac{ 7 }{ \tan 54° } \\ & = \frac{ 7 }{ \text{1.37638...} } \\ & = \text{5.08579...} \\ & \approx \text{5.09} \end{aligned}$

Therefore $x$ is $\text{5.09}$ .

Submit your answer as:

ID is: 3105 Seed is: 9527

Trigonometry with non-right triangles

The figure below, which is drawn to scale, shows an isosceles triangle. Side $\overline{PR} = \text{7.3}$ is labelled and $\hat{P} = \text{57.2}°$ . Point $S$ is on side $\overline{PQ}$ directly across from point $R$ , such that $\overline{RS}$ makes a right angle with $\overline{PQ}$ .

Which of the following statements must be true about the figure above?

Answer: The true statement is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Remember that isosceles triangles are symmetric.

STEP: Consider the choices and select the correct option
[−1 point ⇒ 0 / 1 points left]

The question asks us to select the true statement from the choices in the list. Note that two of the options must be wrong straightaway: $\Delta PQR$ is neither a right-angled triangle nor equilateral. From the remaining options, the correct choice is based on the fact that the segment $\overline{RS}$ divides $\Delta PQR$ into two identical shapes: the segment $\overline{RS}$ is a line of symmetry because the triangle is isosceles. Therefore, point $S$ divides $\overline{PQ}$ into two equal parts.

The correct choice from the list is: $\overline{PS}$ = $\overline{SQ}$ .

Submit your answer as:
Now determine the length of $\overline{PQ}$ . Round your answer to one decimal place.

Answer: The length of $\overline{PQ}$ is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the segment $\overline{RS}$ to get two right-angled triangles. Then you can use trigonometric ratios or the theorem of Pythagoras to work out the answer to the question.

STEP: Draw a line to create right-angled triangles
[−1 point ⇒ 2 / 3 points left]

The first thing to do is draw an extra line across the triangle so that we make two right-angled triangles in the figure. We do this because we can use the trigonometric ratios and the theorem of Pythagoras for right-angled triangles.

The line segment $\overline{SR}$ in the figure is the line we want: it will create two separate right-angled triangles! The two right-angled triangles that we get look like this:

STEP: Use trigonometry to find useful information
[−1 point ⇒ 1 / 3 points left]

In $\Delta PSR$ (the light blue one) we know one of the non-right angles and one of the sides. Hence we can use the trigonometric ratios in that triangle because it is a right-angled triangle.

With the information given, we can find both segments $\overline{PS}$ and $\overline{RS}$ . However, for this question only $\overline{PS}$ is useful: we want the length of $\overline{PQ}$ , which is twice as long as $\overline{PS}$ . In $\Delta PSR$ , the hypotenuse is $\overline{PR} = \text{7.3}$ , and the side we want is adjacent to the angle given. Therefore we need to use the cosine ratio. Set up the equation and then solve for the length of $\overline{PS}$ .

$\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos \left( \text{57.2}° \right) & = \frac{\overline{PS}}{\text{7.3}}\\ \left( \text{7.3} \right) \cos \text{57.2}° & = \overline{PS}\\ \left( \text{7.3} \right) \left(\text{0.5417...}\right) & = \overline{PS}\\ \text{3.9544...} & = \overline{PS} \end{aligned}$

STEP: Calculate the final answer
[−1 point ⇒ 0 / 3 points left]

We want the length of $\overline{PQ}$ , so multiply by two since $S$ is the mid-point of segment $\overline{PQ}$ :

$\begin{aligned} \overline{PQ} & = 2 \cdot \overline{PS}\\ & = 2 \left( \text{3.9544...} \right)\\ & = \text{7.9089...}\\ & \approx \text{7.9} \end{aligned}$

Remember that the instructions say to round the answer to the first decimal place, as shown in the last step above.

The final answer is: $\overline{PQ} = \text{7.9}$ .

Submit your answer as:

ID is: 3105 Seed is: 4936

Trigonometry with non-right triangles

The figure below, which is drawn to scale, shows a scalene triangle. Two sides and an angle are given: $\overline{MP} = \text{8.2}$ , $\; \overline{MN} = \text{10.5}$ and $\hat{M} = \text{34}°$ . Point $Q$ is on side $\overline{MN}$ as labelled, such that $\overline{PQ}$ makes a right angle with $\overline{MN}$ .

Which of the following statements must be true about the figure above?

Answer: The true statement is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Imagine walking from point $M$ to $Q$ and then walking on from $Q$ to $N$ . How does this compare to walking straight from $M$ to $N$ without stopping?

STEP: Consider the choices and select the correct option
[−1 point ⇒ 0 / 1 points left]

The question asks us to select the true statement from the choices in the list. The correct option is about the sum of the segments along side $\overline{MN}$ . Point $Q$ breaks $\overline{MN}$ up into two pieces. Therefore we know that together they make the total length of $\overline{MN}$ . Note that we do not know the exact position of point $Q$ , only that it is somewhere between $M$ and $N$ .

The correct choice from the list is: Side $MN$ is the sum of sides $MQ$ and $QN$ .

Submit your answer as:
Compute the measure of $\hat{N}$ . Round your answer to one decimal place.

Answer: The size of $\hat{N}$ is °.
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

Use the segment $\overline{PQ}$ to get two right-angled triangles. Then you can use trigonometric ratios or the theorem of Pythagoras to work out the answer to the question.

STEP: Draw a line to create right-angled triangles
[−1 point ⇒ 6 / 7 points left]

The first thing to do is draw an extra line across the triangle so that we make two right-angled triangles in the figure. We do this because we can use the trigonometric ratios and the theorem of Pythagoras for right-angled triangles.

The line segment $\overline{QP}$ in the figure is the line we want: it will create two separate right-angled triangles! The two right-angled triangles that we get look like this:

STEP: Use trigonometry to find useful information
[−4 points ⇒ 2 / 7 points left]

In $\Delta MQP$ (the light blue one) we know one of the non-right angles and one of the sides. Hence we can use the trigonometric ratios in that triangle because it is a right-angled triangle.

With the information given, we can find both segments $\overline{MQ}$ and $\overline{PQ}$ . Remember that we want to get the measure of $\hat{N}$ , and for that we need both of these lengths. Start by calculating the length of $\overline{MQ}$ , which will allow us to find $\overline{QN}$ (because we know that $\overline{MN} = \text{10.5}$ ). This calculation involves the hypotenuse and the side adjacent to $\hat{M}$ , so use the cosine ratio.

$\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos \left( \text{34}° \right) & = \frac{\overline{MQ}}{\text{8.2}}\\ \left( \text{8.2} \right) \cos \text{34}° & = \overline{MQ}\\ \left( \text{8.2} \right) \left(\text{0.8290...}\right) & = \overline{MQ}\\ \text{6.7981...} & = \overline{MQ}\\ \text{Since} \quad \overline{MQ} + \overline{QN} = \overline{MN}: &\\ & \overline{QN} = \text{10.5} - \text{6.7981...}\\ & \overline{QN} = \text{3.7018...} \end{aligned}$

Great: that gets us the value for side $\overline{QN}$ . Now we need to find the length of side $\overline{PQ}$ . For that we will use the sine ratio (you can also do this calculation with the theorem of Pythagoras, but here we will do it with trigonometry).

$\begin{aligned}\sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}}\\ \sin \left( \text{34}° \right) & = \frac{\overline{PQ}}{\text{8.2}}\\ \left( \text{8.2} \right) \sin \text{34}° & = \overline{PQ}\\ \left( \text{8.2} \right) \left(\text{0.5591...}\right) & = \overline{PQ}\\ \text{4.5853...} & = \overline{PQ} \end{aligned}$

STEP: Calculate the final answer
[−2 points ⇒ 0 / 7 points left]

Now we can finally work out the angle that we want. $\overline{PQ} = \text{4.5853...}$ and $\overline{QN} = \text{3.7018...}$ are the opposite and adjacent sides for angle $\hat{N}$ , respectively, so this is a tangent ratio situation.

$\begin{aligned} \tan \theta & = \frac{\text{opposite}}{\text{adjacent}}\\ \tan \hat{N} & = \frac{\text{4.5853...}}{\text{3.7018...}}\\ \hat{N} & = \tan^{-1} \left( \frac{\text{4.5853...}}{\text{3.7018...}} \right)\\ & = \text{51.0852...}\\ & \approx \text{51.1}\text{°} \end{aligned}$

Remember that the instructions say to round the answer to the first decimal place, as shown in the last step above.

The final answer is: $\hat{N} = \text{51.1}\text{°}$ .

Submit your answer as:

ID is: 3105 Seed is: 2811

Trigonometry with non-right triangles

The figure below, which is drawn to scale, shows a scalene triangle. Two sides and an angle are given: $\overline{DF} = \text{6.7}$ , $\; \overline{DE} = \text{8}$ and $\hat{D} = \text{35.9}°$ . Point $G$ is on side $\overline{DE}$ as labelled, such that $\overline{FG}$ makes a right angle with $\overline{DE}$ .

Which of the following statements must be true about the figure above?

Answer: The true statement is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Imagine walking from point $D$ to $G$ and then walking on from $G$ to $E$ . How does this compare to walking straight from $D$ to $E$ without stopping?

STEP: Consider the choices and select the correct option
[−1 point ⇒ 0 / 1 points left]

The question asks us to select the true statement from the choices in the list. The correct option is about the sum of the segments along side $\overline{DE}$ . Point $G$ breaks $\overline{DE}$ up into two pieces. Therefore we know that together they make the total length of $\overline{DE}$ . Note that we do not know the exact position of point $G$ , only that it is somewhere between $D$ and $E$ .

The correct choice from the list is: $\overline{DG} + \overline{GE} = \overline{DE}$ .

Submit your answer as:
Next, find the length of $\overline{EF}$ . Round your answer to one decimal place.

Answer: The length of $\overline{EF}$ is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

Use the segment $\overline{FG}$ to get two right-angled triangles. Then you can use trigonometric ratios or the theorem of Pythagoras to work out the answer to the question.

STEP: Draw a line to create right-angled triangles
[−1 point ⇒ 6 / 7 points left]

The first thing to do is draw an extra line across the triangle so that we make two right-angled triangles in the figure. We do this because we can use the trigonometric ratios and the theorem of Pythagoras for right-angled triangles.

The line segment $\overline{GF}$ in the figure is the line we want: it will create two separate right-angled triangles! The two right-angled triangles that we get look like this:

STEP: Use trigonometry to find useful information
[−4 points ⇒ 2 / 7 points left]

In $\Delta DGF$ (the light blue one) we know one of the non-right angles and one of the sides. Hence we can use the trigonometric ratios in that triangle because it is a right-angled triangle.

With the information given, we can find both segments $\overline{DG}$ and $\overline{FG}$ . Remember that we want to get the length of $\overline{EF}$ , and for that we need both of these lengths. Start by calculating the length of $\overline{DG}$ , which will allow us to find $\overline{GE}$ (because we know that $\overline{DE} = \text{8}$ ). This calculation involves the hypotenuse and the side adjacent to $\hat{D}$ , so use the cosine ratio.

$\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos \left( \text{35.9}° \right) & = \frac{\overline{DG}}{\text{6.7}}\\ \left( \text{6.7} \right) \cos \text{35.9}° & = \overline{DG}\\ \left( \text{6.7} \right) \left(\text{0.8100...}\right) & = \overline{DG}\\ \text{5.4272...} & = \overline{DG}\\ \text{Since} \quad \overline{DG} + \overline{GE} = \overline{DE}: &\\ & \overline{GE} = \text{8} - \text{5.4272...}\\ & \overline{GE} = \text{2.5727...} \end{aligned}$

Great: that gets us the value for side $\overline{GE}$ . Now we need to find the length of side $\overline{FG}$ . For that we will use the sine ratio (you can also do this calculation with the theorem of Pythagoras, but here we will do it with trigonometry).

$\begin{aligned}\sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}}\\ \sin \left( \text{35.9}° \right) & = \frac{\overline{FG}}{\text{6.7}}\\ \left( \text{6.7} \right) \sin \text{35.9}° & = \overline{FG}\\ \left( \text{6.7} \right) \left(\text{0.5863...}\right) & = \overline{FG}\\ \text{3.9286...} & = \overline{FG} \end{aligned}$

STEP: Calculate the final answer
[−2 points ⇒ 0 / 7 points left]

Now we have the lengths of two sides of $\Delta GEF$ ; since $\Delta GEF$ is a right-angled triangle, we can use the theorem of Pythagoras to calculate the length of the hypotenuse.

$\begin{aligned} c^{2} & = a^{2} + b^{2}\\ \left( \overline{EF} \right)^{2} & = \left( \overline{FG} \right)^{2} + \left( \overline{GE} \right)^{2}\\ & =\left( \text{3.9286...} \right)^{2} + \left( \text{2.5727...} \right)^{2}\\ & = \text{22.0535...}\\ \therefore \overline{EF} & = \pm \sqrt{\text{22.0535...}} \quad \leftarrow \quad ^{\text{The square root brings in}}_{\text{the } \pm \text{ because } a^{2} = (-a)^{2}}\\ & = \pm \text{4.6961...}\\ & \approx \pm \text{4.7} \end{aligned}$

Remember that the instructions say to round the answer to the first decimal place, as shown in the last step above. Also notice that we get two different answers, one positive and one negative. However, the value we calculated represents a distance, so we must throw out the negative answer.

The final answer is: $\overline{EF} = \text{4.7}$ .

Submit your answer as:

ID is: 896 Seed is: 409

Working with trigonometric equations: sine

Consider the right-angled triangle shown below. One of the sides has a length of $\text{8}$ units and another side has a length of $\text{7}$ units, as labelled. One of the angles is labelled $x$ .

Solve for $x$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

x =

$x=$

°

$\text{°}$

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.

STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

To solve this question we need to write a trigonometric equation. In relation to the angle in the question, the sides are the opposite side and the hypotenuse. So the trigonometric ratio to use is $\sin$ . We can write an equation with the given information and $x$ :

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \sin x & = \frac{7}{8} \end{aligned}

$\begin{aligned} \sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}}\\ \sin x & = \frac{7}{8} \end{aligned}$

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for $x$ . We need to use the inverse sine operation: that is how we get the $x$ out of the sine calculation. As always, we must do the same thing to both sides of the equation.

\begin{aligned} \sin x & = \frac{7}{8} \\ \sin^{- 1} (\sin x) & = \sin^{- 1} (\frac{7}{8}) \\ x & = \sin^{- 1} (\frac{7}{8}) \\ = 61.04497... \\ \approx 61.04 ° \end{aligned}

$\begin{aligned} \sin x & = \frac{7}{8} \\ \sin^{-1} \left( \sin x \right) & = \sin^{-1} \left( \frac{ 7 }{ 8 } \right) \\ x & = \sin^{-1} \left( \frac{ 7 }{ 8 } \right) \\ & = \text{61.04497...} \\ & \approx \text{61.04}\text{°} \end{aligned}$

Therefore $x$ is $\text{61.04}\text{°}$ .

Submit your answer as:

ID is: 896 Seed is: 422

Working with trigonometric equations: sine

Consider the right-angled triangle shown below. One angle is labelled $\text{69} \text{°}$ and one side is $\text{5}$ units long. Another one of the sides is labelled $x$ .

Determine the value of $x$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

x =

$x=$

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.

STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \sin 69 ° & = \frac{x}{5} \end{aligned}

$\begin{aligned} \sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}}\\ \sin 69° & = \frac{x}{5} \end{aligned}$

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for $x$ . Multiply by $\text{5}$ to isolate $x$ . Then evaluate the left side of the equation.

\begin{aligned} \sin 69 ° & = \frac{x}{5} \\ (5) \sin 69 ° & = x \\ (5) (0.93358...) & = x \\ 4.66790... & = x \\ 4.67 & \approx x \end{aligned}

$\begin{aligned} \sin 69° & = \frac{x}{5} \\ \left( 5 \right) \sin 69° & = x \\ \left( 5 \right) \left( \text{0.93358...} \right) & = x \\ \text{4.66790...} & = x \\ \text{4.67} & \approx x \end{aligned}$

Therefore $x$ is $\text{4.67}$ .

Submit your answer as:

ID is: 896 Seed is: 7309

Working with trigonometric equations: sine

Consider the right-angled triangle shown below. One angle is labelled $\text{70} \text{°}$ and one side is $\text{6}$ units long. Another one of the sides is labelled $x$ .

Solve for $x$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

x =

$x=$

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by figuring out which trigonometric ratio to use to solve the question. Then write an equation with that ratio and solve it.

STEP: Choose which trigonometric ratio to use and write an equation
[−1 point ⇒ 2 / 3 points left]

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \sin 70 ° & = \frac{6}{x} \end{aligned}

$\begin{aligned} \sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}}\\ \sin 70° & = \frac{6}{x} \end{aligned}$

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation for $x$ . Rearrange the equation to make $x$ the subject. Then evaluate the right side.

\begin{aligned} \sin 70 ° & = \frac{6}{x} \\ x & = \frac{6}{\sin 70 °} \\ = \frac{6}{0.93969...} \\ = 6.38506... \\ \approx 6.39 \end{aligned}

$\begin{aligned} \sin 70° & = \frac{6}{x} \\ x & = \frac{ 6 }{ \sin 70° } \\ & = \frac{ 6 }{ \text{0.93969...} } \\ & = \text{6.38506...} \\ & \approx \text{6.39} \end{aligned}$

Therefore $x$ is $\text{6.39}$ .

Submit your answer as:

2. Special angles

ID is: 3686 Seed is: 8702

Proportionality in the trigonometric ratios

Right-angled triangle $ACB$ is shown in the diagram below. Side $AC$ has length $5$ , and angle $\hat{A}$ has a measure of $45°$ .

Without using a calculator, determine the length of side $BA$ .

INSTRUCTION: Type sqrt( ) if you need to indicate a square root.

Answer:

B A =

$BA =$ .

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The key to answering this question is the value of the given angle - it is one of the special angles! Use the known values of the trigonometric ratios of the special angles to set up an equation to solve for $BA$ .

STEP: Write a trigonometric equation using the given information
[−1 point ⇒ 2 / 3 points left]

The question asks us to find the length of one of the sides of a triangle, given one side and an angle.

Let's look at where the given information is located in the triangle. The given side $AC$ is adjacent to the given angle. And the unknown side $BA$ is the hypotenuse of the triangle. So we can write the following trigonometric equation:

\begin{matrix} \cos 45 ° = \frac{5}{B A} \end{matrix}

$\begin{array}{c} \cos 45° = \dfrac{5}{BA} \end{array}$

STEP: Write a trigonometric equation using the special angle
[−1 point ⇒ 1 / 3 points left]

We could just solve the above equation using a calculator, but the question says no calculators. So we need some more information before we can find the answer.

Let's take a closer look at the $45°$ angle. Is there anything special about it? Yes - it is one of the special angles! And since we know the trigonometric ratios of the special angles, we can write another equation:

\begin{matrix} \cos 45 ° = \frac{1}{\sqrt{2}} \end{matrix}

$\begin{array}{c} \cos 45° = \dfrac{1}{\sqrt{2}} \end{array}$

Since the left hand side of both of these equations is the same trigonometric ratio of $45°$ , we can construct the following proportion:

\begin{matrix} \frac{5}{B A} = \frac{1}{\sqrt{2}} \end{matrix}

$\begin{array}{c} \dfrac{5}{BA} = \dfrac{1}{\sqrt{2}} \end{array}$

NOTE:

This equation is a proportion. It represents the fact that any triangle with the same angle values will have the same ratio values, no matter what the lengths of the sides are. Here is the original triangle, but with another triangle which has the same angles:

The triangles shown above are similar. And that is where the proportion comes from: similar shapes have proportional sides. The trigonometric ratios are built on similarity. Trigonometry does not care how big a triangle is, it only cares about how big the angles are because that determines the ratio of the sides.

STEP: Solve for the length of the unknown side
[−1 point ⇒ 0 / 3 points left]

Now we can rearrange to find the length of side $BA$ . The value we want is in the denominator, so we can flip both fractions upside down to get it into the numerator.

\begin{aligned} \frac{5}{B A} & = \frac{1}{\sqrt{2}} \\ \frac{B A}{5} & = \sqrt{2} \\ B A & = \sqrt{2} \cdot 5 \\ B A & = 5 \sqrt{2} \end{aligned}

$\begin{aligned} \frac{5}{BA} & = \dfrac{1}{\sqrt{2}} \\ \frac{BA}{5} & = \sqrt{2} \\ BA & = \sqrt{2} \cdot 5 \\ BA & = 5 \sqrt{2} \end{aligned}$

So the final answer is $BA = 5 \sqrt{2}$ .

Submit your answer as:

ID is: 3686 Seed is: 9118

Proportionality in the trigonometric ratios

Right-angled triangle $ZYX$ is shown in the diagram below. Side $XZ$ has length $2 \sqrt{2}$ , and angle $\hat{X}$ has a measure of $45°$ .

Without using a calculator, determine the length of side $YX$ .

INSTRUCTION: Type sqrt( ) if you need to indicate a square root.

Answer:

Y X =

$YX =$ .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

STEP: Write a trigonometric equation using the given information
[−1 point ⇒ 2 / 3 points left]

The question asks us to find the length of one of the sides of a triangle, given one side and an angle.

Let's look at where the given information is located in the triangle. The given side $XZ$ is the hypotenuse of the triangle. And the unknown side $YX$ is adjacent to the angle. So we can write the following trigonometric equation:

\begin{matrix} \cos 45 ° = \frac{Y X}{2 \sqrt{2}} \end{matrix}

$\begin{array}{c} \cos 45° = \dfrac{YX}{2 \sqrt{2}} \end{array}$

STEP: Write a trigonometric equation using the special angle
[−1 point ⇒ 1 / 3 points left]

We could just solve the above equation using a calculator, but the question says no calculators. So we need some more information before we can find the answer.

\begin{matrix} \cos 45 ° = \frac{1}{\sqrt{2}} \end{matrix}

$\begin{array}{c} \cos 45° = \dfrac{1}{\sqrt{2}} \end{array}$

Since the left hand side of both of these equations is the same trigonometric ratio of $45°$ , we can construct the following proportion:

\begin{matrix} \frac{Y X}{2 \sqrt{2}} = \frac{1}{\sqrt{2}} \end{matrix}

$\begin{array}{c} \dfrac{YX}{2 \sqrt{2}} = \dfrac{1}{\sqrt{2}} \end{array}$

NOTE:

STEP: Solve for the length of the unknown side
[−1 point ⇒ 0 / 3 points left]

Now we can rearrange to find the length of side $YX$ .

\begin{aligned} \frac{Y X}{2 \sqrt{2}} & = \frac{1}{\sqrt{2}} \\ Y X & = \frac{1 \cdot 2 \sqrt{2}}{\sqrt{2}} \\ Y X & = 2 \end{aligned}

$\begin{aligned} \frac{YX}{2 \sqrt{2}} & = \dfrac{1}{\sqrt{2}} \\ YX & = \frac{1 \cdot 2 \sqrt{2}}{\sqrt{2}} \\ YX & = 2 \end{aligned}$

So the final answer is $YX = 2$ .

Submit your answer as:

ID is: 3686 Seed is: 9242

Proportionality in the trigonometric ratios

Right-angled triangle $CAB$ is shown in the diagram below. Side $CA$ has length $\sqrt{3}$ , and angle $\hat{B}$ has a measure of $30°$ .

Without using a calculator, determine the length of side $BC$ .

INSTRUCTION: Type sqrt( ) if you need to indicate a square root.

Answer:

B C =

$BC =$ .

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

STEP: Write a trigonometric equation using the given information
[−1 point ⇒ 2 / 3 points left]

The question asks us to find the length of one of the sides of a triangle, given one side and an angle.

Let's look at where the given information is located in the triangle. The given side $CA$ is opposite the given angle. And the unknown side $BC$ is the hypotenuse of the triangle. So we can write the following trigonometric equation:

\begin{matrix} \sin 30 ° = \frac{\sqrt{3}}{B C} \end{matrix}

$\begin{array}{c} \sin 30° = \dfrac{\sqrt{3}}{BC} \end{array}$

STEP: Write a trigonometric equation using the special angle
[−1 point ⇒ 1 / 3 points left]

We could just solve the above equation using a calculator, but the question says no calculators. So we need some more information before we can find the answer.

Let's take a closer look at the $30°$ angle. Is there anything special about it? Yes - it is one of the special angles! And since we know the trigonometric ratios of the special angles, we can write another equation:

\begin{matrix} \sin 30 ° = \frac{1}{2} \end{matrix}

$\begin{array}{c} \sin 30° = \dfrac{1}{2} \end{array}$

Since the left hand side of both of these equations is the same trigonometric ratio of $30°$ , we can construct the following proportion:

\begin{matrix} \frac{\sqrt{3}}{B C} = \frac{1}{2} \end{matrix}

$\begin{array}{c} \dfrac{\sqrt{3}}{BC} = \dfrac{1}{2} \end{array}$

NOTE:

STEP: Solve for the length of the unknown side
[−1 point ⇒ 0 / 3 points left]

Now we can rearrange to find the length of side $BC$ . The value we want is in the denominator, so we can flip both fractions upside down to get it into the numerator.

\begin{aligned} \frac{\sqrt{3}}{B C} & = \frac{1}{2} \\ \frac{B C}{\sqrt{3}} & = 2 \\ B C & = 2 \cdot \sqrt{3} \\ B C & = 2 \sqrt{3} \end{aligned}

$\begin{aligned} \frac{\sqrt{3}}{BC} & = \dfrac{1}{2} \\ \frac{BC}{\sqrt{3}} & = 2 \\ BC & = 2 \cdot \sqrt{3} \\ BC & = 2 \sqrt{3} \end{aligned}$

So the final answer is $BC = 2 \sqrt{3}$ .

Submit your answer as:

ID is: 3545 Seed is: 4195

Special angles and equations

Given the following trigonometric equation:

\begin{matrix} 3 \sin (x + 15 °) = \frac{3}{\sqrt{2}} \end{matrix}

$\begin{array}{c}3\sin (x + 15°) = \frac{3}{\sqrt{2}}\end{array}$

What is the value of $x$ , if $\text{0}\text{°} \leq x + 15° \leq \text{90}\text{°}$ .

Answer:

x =

$x =$

°

$°$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by simplifying and removing any coefficients. What remains on the right hand side? The trigonometric ratio on the left hand side means that you can think of the right hand side as the ratio of two sides of a triangle.

STEP: Cancel common factors
[−1 point ⇒ 3 / 4 points left]

The question asks us to solve for $x$ in the following equation:

\begin{matrix} 3 \sin (x + 15 °) = \frac{3}{\sqrt{2}} \end{matrix}

$\begin{array}{c}3\sin (x + 15°) = \frac{3}{\sqrt{2}}\end{array}$

We need to rearrange and simplify the equation until we reach a point where we know the value of $x$ .

The first step is to remove the coefficient on the left hand side of the equation:

\begin{aligned} 3 \sin (x + 15 °) & = \frac{3}{\sqrt{2}} \\ \frac{1}{3} \cdot 3 \sin (x + 15 °) & = \frac{1}{3} \cdot \frac{3}{\sqrt{2}} \\ \sin (x + 15 °) & = \frac{1}{\sqrt{2}} \end{aligned}

$\begin{aligned} 3\sin (x + 15°) & = \frac{3}{\sqrt{2}} \\ \frac{1}{3} \cdot 3 \sin (x + 15°) & = \frac{1}{3} \cdot \frac{3}{\sqrt{2}} \\ \sin (x + 15°) & = \frac{1}{\sqrt{2}} \end{aligned}$

STEP: Draw the correct special triangle
[−2 points ⇒ 1 / 4 points left]

We now have an equation with a trigonometric expression on the left hand side. This tells us that we can think of the right hand side as the ratio of two sides of a triangle. And we know how to draw a triangle with sides of length $1$ and $\sqrt{2}$ . It is one of the special triangles!

Since the equation contains the function $\sin$ , we need to find the angle opposite the side of length $1$ . This is shown in the diagram below:

We can see from the triangle that the angle we need is the $45°$ . So now we know:

\begin{aligned} \sin (x + 15 °) & = \frac{1}{\sqrt{2}} \\ = \sin 45 ° \end{aligned}

$\begin{aligned} \sin (x + 15°) & = \frac{1}{\sqrt{2}} \\ & = \sin 45° \end{aligned}$

STEP: Solve the resulting equation
[−1 point ⇒ 0 / 4 points left]

Now we have everything we need to solve for $x$ . Since both sides of the equation contain the same trigonometric ratio, we can do the following:

\begin{aligned} \sin (x + 15 °) & = \sin 45 ° \\ x + 15 ° & = 45 ° \end{aligned}

$\begin{aligned} \sin (x + 15°) & = \sin 45° \\ x + 15° & = 45° \end{aligned}$

We can rearrange to solve for $x$ :

\begin{aligned} x + 15 ° & = 45 ° \\ x & = 45 ° - 15 ° \\ = 30 ° \end{aligned}

$\begin{aligned} x + 15° & = 45° \\ x & = 45° - 15° \\ & = 30° \end{aligned}$

TIP: Remember that you can check your answer by substituting this value back in to the original equation. Evaluating the left hand side should give you the value you see on the right hand side.

So the final answer is $x = 30°$ .

Submit your answer as:

ID is: 3545 Seed is: 5530

Special angles and equations

Given the following trigonometric equation:

\begin{matrix} 7 \sin (x - 40 °) = \frac{7}{\sqrt{2}} \end{matrix}

$\begin{array}{c}7\sin (x - 40°) = \frac{7}{\sqrt{2}}\end{array}$

What is the value of $x$ , if $\text{0}\text{°} \leq x - 40° \leq \text{90}\text{°}$ .

Answer:

x =

$x =$

°

$°$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

STEP: Cancel common factors
[−1 point ⇒ 3 / 4 points left]

The question asks us to solve for $x$ in the following equation:

\begin{matrix} 7 \sin (x - 40 °) = \frac{7}{\sqrt{2}} \end{matrix}

$\begin{array}{c}7\sin (x - 40°) = \frac{7}{\sqrt{2}}\end{array}$

We need to rearrange and simplify the equation until we reach a point where we know the value of $x$ .

The first step is to remove the coefficient on the left hand side of the equation:

\begin{aligned} 7 \sin (x - 40 °) & = \frac{7}{\sqrt{2}} \\ \frac{1}{7} \cdot 7 \sin (x - 40 °) & = \frac{1}{7} \cdot \frac{7}{\sqrt{2}} \\ \sin (x - 40 °) & = \frac{1}{\sqrt{2}} \end{aligned}

$\begin{aligned} 7\sin (x - 40°) & = \frac{7}{\sqrt{2}} \\ \frac{1}{7} \cdot 7 \sin (x - 40°) & = \frac{1}{7} \cdot \frac{7}{\sqrt{2}} \\ \sin (x - 40°) & = \frac{1}{\sqrt{2}} \end{aligned}$

STEP: Draw the correct special triangle
[−2 points ⇒ 1 / 4 points left]

Since the equation contains the function $\sin$ , we need to find the angle opposite the side of length $1$ . This is shown in the diagram below:

We can see from the triangle that the angle we need is the $45°$ . So now we know:

\begin{aligned} \sin (x - 40 °) & = \frac{1}{\sqrt{2}} \\ = \sin 45 ° \end{aligned}

$\begin{aligned} \sin (x - 40°) & = \frac{1}{\sqrt{2}} \\ & = \sin 45° \end{aligned}$

STEP: Solve the resulting equation
[−1 point ⇒ 0 / 4 points left]

Now we have everything we need to solve for $x$ . Since both sides of the equation contain the same trigonometric ratio, we can do the following:

\begin{aligned} \sin (x - 40 °) & = \sin 45 ° \\ x - 40 ° & = 45 ° \end{aligned}

$\begin{aligned} \sin (x - 40°) & = \sin 45° \\ x - 40° & = 45° \end{aligned}$

We can rearrange to solve for $x$ :

\begin{aligned} x - 40 ° & = 45 ° \\ x & = 45 ° + 40 ° \\ = 85 ° \end{aligned}

$\begin{aligned} x - 40° & = 45° \\ x & = 45° + 40° \\ & = 85° \end{aligned}$

TIP: Remember that you can check your answer by substituting this value back in to the original equation. Evaluating the left hand side should give you the value you see on the right hand side.

So the final answer is $x = 85°$ .

Submit your answer as:

ID is: 3545 Seed is: 1571

Special angles and equations

Given the following trigonometric equation:

\begin{matrix} 8 \sin (x - 35 °) = 4 \end{matrix}

$\begin{array}{c}8\sin (x - 35°) = 4\end{array}$

What is the value of $x$ , if $\text{0}\text{°} \leq x - 35° \leq \text{90}\text{°}$ .

Answer:

x =

$x =$

°

$°$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

STEP: Cancel common factors
[−1 point ⇒ 3 / 4 points left]

The question asks us to solve for $x$ in the following equation:

\begin{matrix} 8 \sin (x - 35 °) = 4 \end{matrix}

$\begin{array}{c}8\sin (x - 35°) = 4\end{array}$

We need to rearrange and simplify the equation until we reach a point where we know the value of $x$ .

The first step is to remove the coefficient on the left hand side of the equation:

\begin{aligned} 8 \sin (x - 35 °) & = 4 \\ \frac{1}{8} \cdot 8 \sin (x - 35 °) & = \frac{1}{8} \cdot 4 \\ \sin (x - 35 °) & = \frac{1}{2} \end{aligned}

$\begin{aligned} 8\sin (x - 35°) & = 4 \\ \frac{1}{8} \cdot 8 \sin (x - 35°) & = \frac{1}{8} \cdot 4 \\ \sin (x - 35°) & = \frac{1}{2} \end{aligned}$

STEP: Draw the correct special triangle
[−2 points ⇒ 1 / 4 points left]

Since the equation contains the function $\sin$ , we need to find the angle opposite the side of length $1$ . This is shown in the diagram below:

We can see from the triangle that the angle we need is the $30°$ . So now we know:

\begin{aligned} \sin (x - 35 °) & = \frac{1}{2} \\ = \sin 30 ° \end{aligned}

$\begin{aligned} \sin (x - 35°) & = \frac{1}{2} \\ & = \sin 30° \end{aligned}$

STEP: Solve the resulting equation
[−1 point ⇒ 0 / 4 points left]

Now we have everything we need to solve for $x$ . Since both sides of the equation contain the same trigonometric ratio, we can do the following:

\begin{aligned} \sin (x - 35 °) & = \sin 30 ° \\ x - 35 ° & = 30 ° \end{aligned}

$\begin{aligned} \sin (x - 35°) & = \sin 30° \\ x - 35° & = 30° \end{aligned}$

We can rearrange to solve for $x$ :

\begin{aligned} x - 35 ° & = 30 ° \\ x & = 30 ° + 35 ° \\ = 65 ° \end{aligned}

$\begin{aligned} x - 35° & = 30° \\ x & = 30° + 35° \\ & = 65° \end{aligned}$

TIP: Remember that you can check your answer by substituting this value back in to the original equation. Evaluating the left hand side should give you the value you see on the right hand side.

So the final answer is $x = 65°$ .

Submit your answer as:

ID is: 3508 Seed is: 7820

The special angles

In trigonometry, we talk about the "special angles".

What are the values of the special angles? Choose the correct answer from the drop down menu below.

Answer: The special angles are .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you can't remember the values of the special angles, you can look them up in the Everything Maths textbook here.

STEP: Identify the special angles
[−1 point ⇒ 0 / 1 points left]

The question asks us to identify the values of the special angles.

The special angles are the acute angles of the right-angled "special" triangles. These angles are special because we can evaluate the trigonometric ratios of these angles exactly.

There are two special triangles. The $45-45-90$ triangle:

And the $30-60-90$ triangle:

In the first triangle, both of the acute angles are equal to $45°$ . In the second triangle, the acute angles are equal to $30°$ and $60°$ .

So the correct answer is the special angles are $30°$ , $45°$ , and $60°$ .

Submit your answer as:

ID is: 3508 Seed is: 6285

The special angles

In trigonometry, we talk about the "special angles".

What are the values of the special angles? Choose the correct answer from the drop down menu below.

Answer: The special angles are .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you can't remember the values of the special angles, you can look them up in the Everything Maths textbook here.

STEP: Identify the special angles
[−1 point ⇒ 0 / 1 points left]

The question asks us to identify the values of the special angles.

The special angles are the acute angles of the right-angled "special" triangles. These angles are special because we can evaluate the trigonometric ratios of these angles exactly.

There are two special triangles. The $30-60-90$ triangle:

And the $45-45-90$ triangle:

In the first triangle, the acute angles are equal to $30°$ and $60°$ . In the second triangle, both of the acute angles are equal to $45°$ .

So the correct answer is the special angles are $30°$ , $45°$ , and $60°$ .

Submit your answer as:

ID is: 3508 Seed is: 7978

The special angles

In trigonometry, we talk about the "special angles".

What are the values of the special angles? Choose the correct answer from the drop down menu below.

Answer: The special angles are .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you can't remember the values of the special angles, you can look them up in the Everything Maths textbook here.

STEP: Identify the special angles
[−1 point ⇒ 0 / 1 points left]

The question asks us to identify the values of the special angles.

The special angles are the acute angles of the right-angled "special" triangles. These angles are special because we can evaluate the trigonometric ratios of these angles exactly.

There are two special triangles. The $30-60-90$ triangle:

And the $45-45-90$ triangle:

In the first triangle, the acute angles are equal to $30°$ and $60°$ . In the second triangle, both of the acute angles are equal to $45°$ .

So the correct answer is the special angles are $30°$ , $45°$ , and $60°$ .

Submit your answer as:

ID is: 3543 Seed is: 1665

Special angle trigonometric ratios

The table below shows the trigonometric ratios of the special angles. Two of the values are missing from the table. Fill in the missing ratio values A and B.

INSTRUCTION: Do not use a calculator. Type sqrt( ) if you need to type in a square root.

	$30°$	$45°$	$60°$
$\sin$	A	$\dfrac{1}{\sqrt{2}}$	$\dfrac{\sqrt{3}}{2}$
$\cos$	$\dfrac{\sqrt{3}}{2}$	B	$\dfrac{1}{2}$
$\tan$	$\dfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$

Answer:

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If you can't remember the value of the trigonometric ratios of the special angles, start by drawing the special triangles. Then find the correct ratios of the sides.

STEP: Determine the values of the missing ratios
[−2 points ⇒ 0 / 2 points left]

The question asks us to fill in the missing trigonometric ratios of the special angles. The special angles are the acute angles in the special triangles. The angles are special because we can determine the exact values of the trigonometric ratios for these angles.

The special triangles are shown below. The $30-60-90$ triangle is:

And the $45-45-90$ triangle is:

One of the values that we need to find is $\sin 30°$ . Looking at the $30-60-90$ triangle, we can see that the side opposite the $30°$ angle has length $1$ and the hypotenuse has length $2$ . Taking the ratio of these gives us

\begin{matrix} \sin 30 ° = \frac{1}{2} \end{matrix}

$\begin{array}{c}\sin 30° = \dfrac{1}{2}\end{array}$

We can do the same thing to find $\cos 45°$ . Looking at the $45-45-90$ triangle, we can see that the side adjacent to the $45°$ angle has length $1$ and the hypotenuse has length $\sqrt{2}$ . Taking the ratio of these gives us

\begin{matrix} \cos 45 ° = \frac{1}{\sqrt{2}} \end{matrix}

$\begin{array}{c}\cos 45° = \dfrac{1}{\sqrt{2}}\end{array}$

So the completed table looks like this:

	$30°$	$45°$	$60°$
$\sin$	$\dfrac{1}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{\sqrt{3}}{2}$
$\cos$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{2}$
$\tan$	$\dfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$

Submit your answer as: and

ID is: 3543 Seed is: 8874

Special angle trigonometric ratios

The table below shows the trigonometric ratios of the special angles. Two of the values are missing from the table. Fill in the missing ratio values A and B.

INSTRUCTION: Do not use a calculator. Type sqrt( ) if you need to type in a square root.

	$30°$	$45°$	$60°$
$\sin$	A	$\dfrac{1}{\sqrt{2}}$	$\dfrac{\sqrt{3}}{2}$
$\cos$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{2}}$	B
$\tan$	$\dfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$

Answer:

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If you can't remember the value of the trigonometric ratios of the special angles, start by drawing the special triangles. Then find the correct ratios of the sides.

STEP: Determine the values of the missing ratios
[−2 points ⇒ 0 / 2 points left]

The special triangles are shown below. The $30-60-90$ triangle is:

And the $45-45-90$ triangle is:

\begin{matrix} \sin 30 ° = \frac{1}{2} \end{matrix}

$\begin{array}{c}\sin 30° = \dfrac{1}{2}\end{array}$

We can do the same thing to find $\cos 60°$ . Looking at the $30-60-90$ triangle, we can see that the side adjacent to the $60°$ angle has length $1$ and the hypotenuse has length $2$ . Taking the ratio of these gives us

\begin{matrix} \cos 60 ° = \frac{1}{2} \end{matrix}

$\begin{array}{c}\cos 60° = \dfrac{1}{2}\end{array}$

So the completed table looks like this:

	$30°$	$45°$	$60°$
$\sin$	$\dfrac{1}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{\sqrt{3}}{2}$
$\cos$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{2}$
$\tan$	$\dfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$

Submit your answer as: and

ID is: 3543 Seed is: 4845

Special angle trigonometric ratios

The table below shows the trigonometric ratios of the special angles. Two of the values are missing from the table. Fill in the missing ratio values A and B.

INSTRUCTION: Do not use a calculator. Type sqrt( ) if you need to type in a square root.

	$30°$	$45°$	$60°$
$\sin$	A	$\dfrac{1}{\sqrt{2}}$	$\dfrac{\sqrt{3}}{2}$
$\cos$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{2}$
$\tan$	$\dfrac{1}{\sqrt{3}}$	B	$\sqrt{3}$

Answer:

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If you can't remember the value of the trigonometric ratios of the special angles, start by drawing the special triangles. Then find the correct ratios of the sides.

STEP: Determine the values of the missing ratios
[−2 points ⇒ 0 / 2 points left]

The special triangles are shown below. The $30-60-90$ triangle is:

And the $45-45-90$ triangle is:

\begin{matrix} \sin 30 ° = \frac{1}{2} \end{matrix}

$\begin{array}{c}\sin 30° = \dfrac{1}{2}\end{array}$

We can do the same thing to find $\tan 45°$ . Looking at the $45-45-90$ triangle, we can see that the side opposite the $45°$ angle has length $1$ and the side adjacent to it has length $1$ . Taking the ratio of these gives us

\begin{matrix} \tan 45 ° = 1 \end{matrix}

$\begin{array}{c}\tan 45° = 1\end{array}$

So the completed table looks like this:

	$30°$	$45°$	$60°$
$\sin$	$\dfrac{1}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{\sqrt{3}}{2}$
$\cos$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{2}$
$\tan$	$\dfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$

Submit your answer as: and

ID is: 3522 Seed is: 3277

The special angles

In trigonometry, we talk about the "special angles". What is the value of one of these special angles?

Your answer should be a number greater than zero and less than $90°$ .

Answer: One of the special angles is

°

$°$ .

one-of

type(numeric.noerror)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

There are three different special angles. These are the acute angles in the special triangles. If you can't remember what the special triangles look like, you can see them in the Everything Maths textbook here.

STEP: Remember the values of the special angles
[−1 point ⇒ 0 / 1 points left]

The special angles are the acute angles in the special triangles. The angles are special because we can determine the exact values of the trigonometric ratios for these angles.

The special triangles are shown below.

TIP: You need to remember how to draw these triangles for tests and exams.

These diagrams show that the values of the acute angles in the special triangles are $30°$ , $45°$ , and $60°$ . You could have given any one of these values as your answer.

Any one of these three angles is acceptable: $30°$ , $45°$ , or $60°$ .

Submit your answer as:

ID is: 3522 Seed is: 2826

The special angles

In trigonometry, we talk about the "special angles". What is the value of one of these special angles?

Your answer should be a number greater than zero and less than $90°$ .

Answer: One of the special angles is

°

$°$ .

one-of

type(numeric.noerror)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

STEP: Remember the values of the special angles
[−1 point ⇒ 0 / 1 points left]

The special angles are the acute angles in the special triangles. The angles are special because we can determine the exact values of the trigonometric ratios for these angles.

The special triangles are shown below.

TIP: You need to remember how to draw these triangles for tests and exams.

These diagrams show that the values of the acute angles in the special triangles are $30°$ , $45°$ , and $60°$ . You could have given any one of these values as your answer.

Any one of these three angles is acceptable: $30°$ , $45°$ , or $60°$ .

Submit your answer as:

ID is: 3522 Seed is: 8424

The special angles

In trigonometry, we talk about the "special angles". What is the value of one of these special angles?

Your answer should be a number greater than zero and less than $90°$ .

Answer: One of the special angles is

°

$°$ .

one-of

type(numeric.noerror)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

STEP: Remember the values of the special angles
[−1 point ⇒ 0 / 1 points left]

The special angles are the acute angles in the special triangles. The angles are special because we can determine the exact values of the trigonometric ratios for these angles.

The special triangles are shown below.

TIP: You need to remember how to draw these triangles for tests and exams.

These diagrams show that the values of the acute angles in the special triangles are $30°$ , $45°$ , and $60°$ . You could have given any one of these values as your answer.

Any one of these three angles is acceptable: $30°$ , $45°$ , or $60°$ .

Submit your answer as:

ID is: 3544 Seed is: 2992

Special angles: what is the angle?

Given the following trigonometric equation:

\begin{matrix} \tan x = 1 \end{matrix}

$\begin{array}{c}\tan x = 1\end{array}$

What is the value of $x$ ?

Answer:

x =

$x =$

°

$°$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The left hand side of the equation is a trigonometric expression. This means that you can think of the right hand side as the ratio of two sides of a triangle. We can rewrite the $1$ as $\dfrac{1}{1}$ . Then we have information about two of the sides. What triangle do you know that has sides of length $1$ and $1$ ? The answer will be one of the angles of this triangle.

STEP: Draw a special triangle
[−1 point ⇒ 1 / 2 points left]

The question asks us to solve for $x$ in the following equation:

\begin{matrix} \tan x = 1 \end{matrix}

$\begin{array}{c}\tan x = 1\end{array}$

We could solve this using a calculator. But a closer look at the equation shows us that we don't need to.

The left hand side of the equation is a trigonometric ratio. This tells us that we can think of the right hand side as the ratio of two sides of a triangle. We can rewrite the $1$ as $\dfrac{1}{1}$ . Then we have information about two of the sides. And we know how to draw a triangle with two sides of length $1$ and $1$ . It is one of the special triangles!

STEP: Solve the equation
[−1 point ⇒ 0 / 2 points left]

Since the equation contains the ratio $\tan$ , we need to find the angle opposite the side of length $1$ and adjacent to the side of length $1$ . This is shown in the diagram below.

We can see from the triangle that the angle we need is the $45°$ angle.

So the answer is $x = 45°$ .

Submit your answer as:

ID is: 3544 Seed is: 7

Special angles: what is the angle?

Given the following trigonometric equation:

\begin{matrix} \sin x = \frac{1}{2} \end{matrix}

$\begin{array}{c}\sin x = \frac{1}{2}\end{array}$

What is the value of $x$ ?

Answer:

x =

$x =$

°

$°$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The left hand side of the equation is a trigonometric expression. This means that you can think of the right hand side as the ratio of two sides of a triangle. What triangle do you know that has sides of length $1$ and $2$ ? The answer will be one of the angles of this triangle.

STEP: Draw a special triangle
[−1 point ⇒ 1 / 2 points left]

The question asks us to solve for $x$ in the following equation:

\begin{matrix} \sin x = \frac{1}{2} \end{matrix}

$\begin{array}{c}\sin x = \frac{1}{2}\end{array}$

We could solve this using a calculator. But a closer look at the equation shows us that we don't need to.

The left hand side of the equation is a trigonometric ratio. This tells us that we can think of the right hand side as the ratio of two sides of a triangle. And we know how to draw a triangle with two sides of length $1$ and $2$ . It is one of the special triangles!

STEP: Solve the equation
[−1 point ⇒ 0 / 2 points left]

Since the equation contains the ratio $\sin$ , we need to find the angle opposite the side of length $1$ . This is shown in the diagram below.

We can see from the triangle that the angle we need is the $30°$ angle.

So the answer is $x = 30°$ .

Submit your answer as:

ID is: 3544 Seed is: 6255

Special angles: what is the angle?

Given the following trigonometric equation:

\begin{matrix} \sin x = \frac{1}{2} \end{matrix}

$\begin{array}{c}\sin x = \frac{1}{2}\end{array}$

What is the value of $x$ ?

Answer:

x =

$x =$

°

$°$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: Draw a special triangle
[−1 point ⇒ 1 / 2 points left]

The question asks us to solve for $x$ in the following equation:

\begin{matrix} \sin x = \frac{1}{2} \end{matrix}

$\begin{array}{c}\sin x = \frac{1}{2}\end{array}$

We could solve this using a calculator. But a closer look at the equation shows us that we don't need to.

STEP: Solve the equation
[−1 point ⇒ 0 / 2 points left]

Since the equation contains the ratio $\sin$ , we need to find the angle opposite the side of length $1$ . This is shown in the diagram below.

We can see from the triangle that the angle we need is the $30°$ angle.

So the answer is $x = 30°$ .

Submit your answer as:

ID is: 3507 Seed is: 954

Identifying special angles

In trigonometry, we talk about the "special angles".

Which one of the following is a special angle? Choose the correct answer from the drop-down menu below.

Answer: One of the special angles is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you can't remember the values of the special angles, you can look them up in the Everything Maths textbook here.

STEP: Identify the special angle
[−1 point ⇒ 0 / 1 points left]

The question asks us to identify the value of one of the special angles.

The special angles are the acute angles in the "special" triangles. What makes them special is the fact that we can evaluate the trigonometric ratios of these angles exactly.

There are two special triangles. The $30-60-90$ triangle:

And the $45-45-90$ triangle:

In the first triangle, the acute angles are equal to $30°$ and $60°$ . In the second triangle, both of the acute angles are equal to $45°$ .

Only one of these special angles was in the list of possible answers: $30°$ .

So the value of one of the special angles is $30°$ .

Submit your answer as:

ID is: 3507 Seed is: 2855

Identifying special angles

In trigonometry, we talk about the "special angles".

Which one of the following is a special angle? Choose the correct answer from the drop-down menu below.

Answer: One of the special angles is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you can't remember the values of the special angles, you can look them up in the Everything Maths textbook here.

STEP: Identify the special angle
[−1 point ⇒ 0 / 1 points left]

The question asks us to identify the value of one of the special angles.

The special angles are the acute angles in the "special" triangles. What makes them special is the fact that we can evaluate the trigonometric ratios of these angles exactly.

There are two special triangles. The $45-45-90$ triangle:

And the $30-60-90$ triangle:

In the first triangle, both of the acute angles are equal to $45°$ . In the second triangle, the acute angles are equal to $30°$ and $60°$ .

Only one of these special angles was in the list of possible answers: $45°$ .

So the value of one of the special angles is $45°$ .

Submit your answer as:

ID is: 3507 Seed is: 2116

Identifying special angles

In trigonometry, we talk about the "special angles".

Which one of the following is a special angle? Choose the correct answer from the drop-down menu below.

Answer: One of the special angles is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you can't remember the values of the special angles, you can look them up in the Everything Maths textbook here.

STEP: Identify the special angle
[−1 point ⇒ 0 / 1 points left]

The question asks us to identify the value of one of the special angles.

The special angles are the acute angles in the "special" triangles. What makes them special is the fact that we can evaluate the trigonometric ratios of these angles exactly.

There are two special triangles. The $30-60-90$ triangle:

And the $45-45-90$ triangle:

In the first triangle, the acute angles are equal to $30°$ and $60°$ . In the second triangle, both of the acute angles are equal to $45°$ .

Only one of these special angles was in the list of possible answers: $60°$ .

So the value of one of the special angles is $60°$ .

Submit your answer as:

ID is: 3520 Seed is: 1473

Memorise those special angle ratios!

The following expression is a trigonometric ratio of one of the special angles.

\begin{matrix} \frac{1}{\sqrt{2}} \end{matrix}

$\begin{array}{c}\dfrac{1}{\sqrt{2}}\end{array}$

Give a trigonometric expression that is equal to this value.

INSTRUCTION: There may be more than one correct answer. Give one correct answer only.

Answer:

\frac{1}{\sqrt{2}} =

$\dfrac{1}{\sqrt{2}} =$

one-of

type(expression)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by drawing the special triangle that has sides of length $1$ and $\sqrt{2}$ . Identify which sides are opposite and adjacent to each of the acute angles. Then find which trigonometric ratio you need to use to get the expression given in the question.

STEP: Draw the correct special triangle
[−1 point ⇒ 1 / 2 points left]

The question asks us to find a trigonometric ratio of a special angle that is equivalent to

\begin{matrix} \frac{1}{\sqrt{2}} \end{matrix}

$\begin{array}{c}\dfrac{1}{\sqrt{2}}\end{array}$

When a question asks us to work with a special angle, it is always a good first step to draw the corresponding special triangle. We do not yet know which angle we need, but we do know something about the sides of the triangle.

We know that one of the sides has length $1$ and another has length $\sqrt{2}$ . This is enough to tell us that we need to draw the following special triangle:

So now we know that the special triangle that we need for this question has acute angles equal to $45°$ .

STEP: Identify a corresponding trigonometric ratio
[−1 point ⇒ 0 / 2 points left]

Now we need to identify a trigonometric ratio of one of the special angles that will give us the fraction given in the question.

Consider the angle $45°$ . The side with length $1$ is opposite $45°$ , and the side with length $\sqrt{2}$ is the hypotenuse (it is always opposite the right-angle). This is shown in the diagram below:

Since $\sin$ is defined as the ratio of the opposite and the hypotenuse, we can see that

\begin{matrix} \frac{1}{\sqrt{2}} = \sin 45 ° \end{matrix}

$\begin{array}{c}\dfrac{1}{\sqrt{2}} = \sin 45°\end{array}$

NOTE: The side of length

1

$1$ is also adjacent to the

45 °

$45°$ angle, so

\frac{1}{\sqrt{2}} = \cos 45 °

$\dfrac{1}{\sqrt{2}} = \cos 45°$ is also a valid solution.

So the answer is $\dfrac{1}{\sqrt{2}} = \sin 45°$ .

Submit your answer as:

ID is: 3520 Seed is: 2071

Memorise those special angle ratios!

The following expression is a trigonometric ratio of one of the special angles.

\begin{matrix} \frac{1}{\sqrt{2}} \end{matrix}

$\begin{array}{c}\dfrac{1}{\sqrt{2}}\end{array}$

Give a trigonometric expression that is equal to this value.

INSTRUCTION: There may be more than one correct answer. Give one correct answer only.

Answer:

\frac{1}{\sqrt{2}} =

$\dfrac{1}{\sqrt{2}} =$

one-of

type(expression)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: Draw the correct special triangle
[−1 point ⇒ 1 / 2 points left]

The question asks us to find a trigonometric ratio of a special angle that is equivalent to

\begin{matrix} \frac{1}{\sqrt{2}} \end{matrix}

$\begin{array}{c}\dfrac{1}{\sqrt{2}}\end{array}$

We know that one of the sides has length $1$ and another has length $\sqrt{2}$ . This is enough to tell us that we need to draw the following special triangle:

So now we know that the special triangle that we need for this question has acute angles equal to $45°$ .

STEP: Identify a corresponding trigonometric ratio
[−1 point ⇒ 0 / 2 points left]

Now we need to identify a trigonometric ratio of one of the special angles that will give us the fraction given in the question.

Since $\sin$ is defined as the ratio of the opposite and the hypotenuse, we can see that

\begin{matrix} \frac{1}{\sqrt{2}} = \sin 45 ° \end{matrix}

$\begin{array}{c}\dfrac{1}{\sqrt{2}} = \sin 45°\end{array}$

NOTE: The side of length

1

$1$ is also adjacent to the

45 °

$45°$ angle, so

\frac{1}{\sqrt{2}} = \cos 45 °

$\dfrac{1}{\sqrt{2}} = \cos 45°$ is also a valid solution.

So the answer is $\dfrac{1}{\sqrt{2}} = \sin 45°$ .

Submit your answer as:

ID is: 3520 Seed is: 9672

Memorise those special angle ratios!

The following expression is a trigonometric ratio of one of the special angles.

\begin{matrix} \sqrt{3} \end{matrix}

$\begin{array}{c}\sqrt{3}\end{array}$

Give a trigonometric expression that is equal to this value.

INSTRUCTION: There may be more than one correct answer. Give one correct answer only.

Answer:

\sqrt{3} =

$\sqrt{3} =$

one-of

type(expression)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

We can rewrite the $\sqrt{3}$ as $\dfrac{\sqrt{3}}{1}$ . Then we have information about two of the sides. Start by drawing the special triangle that has sides of length $\sqrt{3}$ and $1$ . Identify which sides are opposite and adjacent to each of the acute angles. Then find which trigonometric ratio you need to use to get the expression given in the question.

STEP: Draw the correct special triangle
[−1 point ⇒ 1 / 2 points left]

The question asks us to find a trigonometric ratio of a special angle that is equivalent to

\begin{matrix} \sqrt{3} \end{matrix}

$\begin{array}{c}\sqrt{3}\end{array}$

We can rewrite the $\sqrt{3}$ as $\dfrac{\sqrt{3}}{1}$ . Then we have information about two of the sides. We know that one of the sides has length $\sqrt{3}$ and another has length $1$ . This is enough to tell us that we need to draw the following special triangle:

So now we know that the special triangle that we need for this question has acute angles equal to $30°$ and $60°$ .

STEP: Identify a corresponding trigonometric ratio
[−1 point ⇒ 0 / 2 points left]

Now we need to identify a trigonometric ratio of one of the special angles that will give us the fraction given in the question.

Consider the angle $60°$ . The side with length $\sqrt{3}$ is opposite $60°$ , and the side with length $1$ is adjacent to $60°$ . This is shown in the diagram below:

Since $\tan$ is defined as the ratio of the opposite and the adjacent, we can see that

\begin{matrix} \sqrt{3} = \tan 60 ° \end{matrix}

$\begin{array}{c}\sqrt{3} = \tan 60°\end{array}$

So the answer is $\sqrt{3} = \tan 60°$ .

Submit your answer as:

ID is: 3675 Seed is: 2362

Using the special angles to solve problems

Quadrilateral $KLMN$ , constructed from two triangles, is shown in the diagram below. Two angles are labelled: $\angle MKL = 30°$ and $\angle KMN = 45°$ . Side $KL$ has length $\sqrt{6}$ .

Find the length of side $MN$ , marked with a $?$ in the diagram.

INSTRUCTION: Your answer should be exact. Type sqrt( ) if you need to indicate a square root.

Answer: The length of

M N =

$MN =$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by finding the length of the shared side $KM$ . Since $\angle MKL = 30°$ is one of the special angles, you can do this using proportions of the known trigonometric ratios of the special angles.

STEP: Find the length of side

K M

$KM$
[−1 point ⇒ 1 / 2 points left]

The question asks us to find the length of side $MN$ in the diagram, given the values of two angles and one side in the shape.

Since the side we are given and the side we are looking for are not in the same triangle, it will be useful to first find the length of the shared side of the two triangles.

In triangle $KLM$ , we have angle $\angle MKL$ and side $KL$ . We first want to find the length of side $KM$ . So we can use the cosine ratio to write this equation:

\begin{matrix} \cos 30 ° = \frac{\sqrt{6}}{K M} \end{matrix}

$\begin{array}{c} \cos 30° = \dfrac{\sqrt{6}}{KM} \end{array}$

Let's take a closer look at the $30°$ angle. Is there anything special about it? Yes: it is one of the special angles! And since we know the trigonometric ratios of the special angles, we can write another equation:

\begin{matrix} \cos 30 ° = \frac{\sqrt{3}}{2} \end{matrix}

$\begin{array}{c} \cos 30° = \dfrac{\sqrt{3}}{2} \end{array}$

Since the left hand sides of both of these equations are identical, we can solve for the length of side $KM$ :

\begin{aligned} \frac{\sqrt{6}}{K M} & = \frac{\sqrt{3}}{2} \\ K M & = \frac{\sqrt{6} \cdot 2}{\sqrt{3}} \\ K M & = 2 \sqrt{2} \end{aligned}

$\begin{aligned} \frac{\sqrt{6}}{KM} & = \dfrac{\sqrt{3}}{2} \\ KM & = \dfrac{\sqrt{6} \cdot 2}{ \sqrt{3} } \\ KM & = 2 \sqrt{2} \end{aligned}$

Now that we have some more information about the quadrilateral, let's add it to the diagram:

STEP: Find the length of side

M N

$MN$
[−1 point ⇒ 0 / 2 points left]

Now we know one side and an angle in triangle $KNM$ . So we can use trigonometry to find the length of side $MN$ .

We can write the following trigonometric expression for triangle $KNM$ :

\begin{matrix} \cos 45 ° = \frac{2 \sqrt{2}}{M N} \end{matrix}

$\begin{array}{c} \cos 45° = \dfrac{2 \sqrt{2}}{MN} \end{array}$

But $45°$ is also one the special angles! So we can use the same method of proportions to find side $MN$ that we used to find side $KM$ . We first write another equation for $\cos 45°$ :

\begin{matrix} \cos 45 ° = \frac{1}{\sqrt{2}} \end{matrix}

$\begin{array}{c} \cos 45° = \dfrac{1}{\sqrt{2}} \end{array}$

And then use this to solve for $MN$ :

\begin{aligned} \frac{2 \sqrt{2}}{M N} & = \frac{1}{\sqrt{2}} \\ M N & = 2 \sqrt{2} \cdot \sqrt{2} \\ M N & = 4 \end{aligned}

$\begin{aligned} \frac{2 \sqrt{2}}{MN} & = \dfrac{1}{\sqrt{2}} \\ MN & = 2 \sqrt{2} \cdot \sqrt{2} \\ MN & = 4 \end{aligned}$

So the correct answer is side $MN = 4$ .

Submit your answer as:

ID is: 3675 Seed is: 8542

Using the special angles to solve problems

Quadrilateral $KLMN$ , constructed from two triangles, is shown in the diagram below. Two angles are labelled: $\angle KMN = 30°$ and $\angle MKL = 45°$ . Side $NK$ has length $\sqrt{2}$ .

Find the length of side $KL$ , marked with a $?$ in the diagram.

INSTRUCTION: Your answer should be exact. Type sqrt( ) if you need to indicate a square root.

Answer: The length of

K L =

$KL =$

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by finding the length of the shared side $KM$ . Since $\angle KMN = 30°$ is one of the special angles, you can do this using proportions of the known trigonometric ratios of the special angles.

STEP: Find the length of side

K M

$KM$
[−1 point ⇒ 1 / 2 points left]

The question asks us to find the length of side $KL$ in the diagram, given the values of two angles and one side in the shape.

Since the side we are given and the side we are looking for are not in the same triangle, it will be useful to first find the length of the shared side of the two triangles.

In triangle $KNM$ , we have angle $\angle KMN$ and side $NK$ . We first want to find the length of side $KM$ . So we can use the tangent ratio to write this equation:

\begin{matrix} \tan 30 ° = \frac{\sqrt{2}}{K M} \end{matrix}

$\begin{array}{c} \tan 30° = \dfrac{\sqrt{2}}{KM} \end{array}$

\begin{matrix} \tan 30 ° = \frac{1}{\sqrt{3}} \end{matrix}

$\begin{array}{c} \tan 30° = \dfrac{1}{\sqrt{3}} \end{array}$

Since the left hand sides of both of these equations are identical, we can solve for the length of side $KM$ :

\begin{aligned} \frac{\sqrt{2}}{K M} & = \frac{1}{\sqrt{3}} \\ K M & = \sqrt{2} \cdot \sqrt{3} \\ K M & = \sqrt{6} \end{aligned}

$\begin{aligned} \frac{\sqrt{2}}{KM} & = \dfrac{1}{\sqrt{3}} \\ KM & = \sqrt{2} \cdot \sqrt{3} \\ KM & = \sqrt{6} \end{aligned}$

Now that we have some more information about the quadrilateral, let's add it to the diagram:

STEP: Find the length of side

K L

$KL$
[−1 point ⇒ 0 / 2 points left]

Now we know one side and an angle in triangle $KLM$ . So we can use trigonometry to find the length of side $KL$ .

We can write the following trigonometric expression for triangle $KLM$ :

\begin{matrix} \cos 45 ° = \frac{K L}{\sqrt{6}} \end{matrix}

$\begin{array}{c} \cos 45° = \dfrac{KL}{\sqrt{6}} \end{array}$

But $45°$ is also one the special angles! So we can use the same method of proportions to find side $KL$ that we used to find side $KM$ . We first write another equation for $\cos 45°$ :

\begin{matrix} \cos 45 ° = \frac{1}{\sqrt{2}} \end{matrix}

$\begin{array}{c} \cos 45° = \dfrac{1}{\sqrt{2}} \end{array}$

And then use this to solve for $KL$ :

\begin{aligned} \frac{K L}{\sqrt{6}} & = \frac{1}{\sqrt{2}} \\ K L & = \frac{\sqrt{6} \cdot 1}{\sqrt{2}} \\ K L & = \sqrt{3} \end{aligned}

$\begin{aligned} \frac{KL}{\sqrt{6}} & = \dfrac{1}{\sqrt{2}} \\ KL & = \dfrac{\sqrt{6} \cdot 1}{ \sqrt{2} } \\ KL & = \sqrt{3} \end{aligned}$

So the correct answer is side $KL = \sqrt{3}$ .

Submit your answer as:

ID is: 3675 Seed is: 7002

Using the special angles to solve problems

Quadrilateral $KLMN$ , constructed from two triangles, is shown in the diagram below. Two angles are labelled: $\angle NKM = 45°$ and $\angle MKL = 60°$ . Side $NK$ has length $\sqrt{6}$ .

Find the length of side $LM$ , marked with a $?$ in the diagram.

INSTRUCTION: Your answer should be exact. Type sqrt( ) if you need to indicate a square root.

Answer: The length of

L M =

$LM =$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by finding the length of the shared side $KM$ . Since $\angle NKM = 45°$ is one of the special angles, you can do this using proportions of the known trigonometric ratios of the special angles.

STEP: Find the length of side

K M

$KM$
[−1 point ⇒ 1 / 2 points left]

The question asks us to find the length of side $LM$ in the diagram, given the values of two angles and one side in the shape.

Since the side we are given and the side we are looking for are not in the same triangle, it will be useful to first find the length of the shared side of the two triangles.

In triangle $KNM$ , we have angle $\angle NKM$ and side $NK$ . We first want to find the length of side $KM$ . So we can use the cosine ratio to write this equation:

\begin{matrix} \cos 45 ° = \frac{K M}{\sqrt{6}} \end{matrix}

$\begin{array}{c} \cos 45° = \dfrac{KM}{\sqrt{6}} \end{array}$

Let's take a closer look at the $45°$ angle. Is there anything special about it? Yes: it is one of the special angles! And since we know the trigonometric ratios of the special angles, we can write another equation:

\begin{matrix} \cos 45 ° = \frac{1}{\sqrt{2}} \end{matrix}

$\begin{array}{c} \cos 45° = \dfrac{1}{\sqrt{2}} \end{array}$

Since the left hand sides of both of these equations are identical, we can solve for the length of side $KM$ :

\begin{aligned} \frac{K M}{\sqrt{6}} & = \frac{1}{\sqrt{2}} \\ K M & = \frac{\sqrt{6} \cdot 1}{\sqrt{2}} \\ K M & = \sqrt{3} \end{aligned}

$\begin{aligned} \frac{KM}{\sqrt{6}} & = \dfrac{1}{\sqrt{2}} \\ KM & = \dfrac{\sqrt{6} \cdot 1}{ \sqrt{2} } \\ KM & = \sqrt{3} \end{aligned}$

Now that we have some more information about the quadrilateral, let's add it to the diagram:

STEP: Find the length of side

L M

$LM$
[−1 point ⇒ 0 / 2 points left]

Now we know one side and an angle in triangle $KLM$ . So we can use trigonometry to find the length of side $LM$ .

We can write the following trigonometric expression for triangle $KLM$ :

\begin{matrix} \sin 60 ° = \frac{L M}{\sqrt{3}} \end{matrix}

$\begin{array}{c} \sin 60° = \dfrac{LM}{\sqrt{3}} \end{array}$

But $60°$ is also one the special angles! So we can use the same method of proportions to find side $LM$ that we used to find side $KM$ . We first write another equation for $\sin 60°$ :

\begin{matrix} \sin 60 ° = \frac{\sqrt{3}}{2} \end{matrix}

$\begin{array}{c} \sin 60° = \dfrac{\sqrt{3}}{2} \end{array}$

And then use this to solve for $LM$ :

\begin{aligned} \frac{L M}{\sqrt{3}} & = \frac{\sqrt{3}}{2} \\ L M & = \frac{\sqrt{3} \cdot \sqrt{3}}{2} \\ L M & = \frac{3}{2} \end{aligned}

$\begin{aligned} \frac{LM}{\sqrt{3}} & = \dfrac{\sqrt{3}}{2} \\ LM & = \dfrac{\sqrt{3} \cdot \sqrt{3}}{ 2 } \\ LM & = \frac{3}{2} \end{aligned}$

So the correct answer is side $LM = \dfrac{3}{2}$ .

Submit your answer as:

ID is: 919 Seed is: 3506

Trigonometry: special angles

Evaluate the following trigonometric expression:

\begin{matrix} \tan 30 ° \end{matrix}

$\begin{array}{c}\tan30°\end{array}$

INSTRUCTION: Do not use a calculator.

Select the answer from the table below:

A	B	C	D	E
$\dfrac{1}{2}$	$\dfrac{\sqrt{3}}{2}$	$1$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{\sqrt{3}}$

Answer: The correct answer is option .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The expression contains the angle $30°$ , which is a special angle! This means that we can evaluate the ratio $\tan30°$ exactly. If you can't remember the value of $\tan30°$ , start by drawing the special triangle which has the angle $30°$ in it. Then use the lengths of the sides of the special triangle to evaluate the ratio $\tan30°$ .

STEP: Draw a special triangle and use it to evaluate the expression

\tan 30 °

$\tan30°$
[−1 point ⇒ 0 / 1 points left]

The question asks us to evaluate the trigonometric expression

\begin{matrix} \tan 30 ° \end{matrix}

$\begin{array}{c}\tan30°\end{array}$

One way to find the answer would be to type this into a calculator. But the question specifically says no calculators!

Lucky for us, $30°$ is a special angle! Special angles are angles for which we can evaluate the trigonometric ratios exactly.

We can do this using the special triangles. The special triangle that contains the angle $30°$ is shown below. Since we want to find $\tan30°$ , we need to identify the opposite and the adjacent.

Taking the ratio of the opposite and the adjacent, we can see that

\begin{matrix} \tan 30 ° = \frac{1}{\sqrt{3}} \end{matrix}

$\begin{array}{c}\tan30° = \dfrac{1}{\sqrt{3}}\end{array}$

NOTE: You need to know the trigonometric ratios of the special angles for tests and exams. If you can't remember all of the ratios, start by drawing the correct special triangle. Then evaluate the ratio using the picture.

The correct answer is E: $\tan30° = \dfrac{1}{\sqrt{3}}$ .

Submit your answer as:

ID is: 919 Seed is: 6033

Trigonometry: special angles

Evaluate the following trigonometric expression:

\begin{matrix} \sin 60 ° \end{matrix}

$\begin{array}{c}\sin60°\end{array}$

INSTRUCTION: Do not use a calculator.

Select the answer from the table below:

A	B	C	D	E
$\dfrac{1}{\sqrt{2}}$	$\dfrac{\sqrt{3}}{2}$	$\sqrt{3}$	$\dfrac{1}{2}$	$1$

Answer: The correct answer is option .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The expression contains the angle $60°$ , which is a special angle! This means that we can evaluate the ratio $\sin60°$ exactly. If you can't remember the value of $\sin60°$ , start by drawing the special triangle which has the angle $60°$ in it. Then use the lengths of the sides of the special triangle to evaluate the ratio $\sin60°$ .

STEP: Draw a special triangle and use it to evaluate the expression

\sin 60 °

$\sin60°$
[−1 point ⇒ 0 / 1 points left]

The question asks us to evaluate the trigonometric expression

\begin{matrix} \sin 60 ° \end{matrix}

$\begin{array}{c}\sin60°\end{array}$

One way to find the answer would be to type this into a calculator. But the question specifically says no calculators!

Lucky for us, $60°$ is a special angle! Special angles are angles for which we can evaluate the trigonometric ratios exactly.

We can do this using the special triangles. The special triangle that contains the angle $60°$ is shown below. Since we want to find $\sin60°$ , we need to identify the opposite and the hypotenuse.

Taking the ratio of the opposite and the hypotenuse, we can see that

\begin{matrix} \sin 60 ° = \frac{\sqrt{3}}{2} \end{matrix}

$\begin{array}{c}\sin60° = \dfrac{\sqrt{3}}{2}\end{array}$

The correct answer is B: $\sin60° = \dfrac{\sqrt{3}}{2}$ .

Submit your answer as:

ID is: 919 Seed is: 4623

Trigonometry: special angles

Evaluate the following trigonometric expression:

\begin{matrix} \sin 30 ° \end{matrix}

$\begin{array}{c}\sin30°\end{array}$

INSTRUCTION: Do not use a calculator.

Select the answer from the table below:

A	B	C	D	E
$\dfrac{1}{\sqrt{2}}$	$\sqrt{3}$	$\dfrac{1}{\sqrt{3}}$	$\dfrac{1}{2}$	$\dfrac{\sqrt{3}}{2}$

Answer: The correct answer is option .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The expression contains the angle $30°$ , which is a special angle! This means that we can evaluate the ratio $\sin30°$ exactly. If you can't remember the value of $\sin30°$ , start by drawing the special triangle which has the angle $30°$ in it. Then use the lengths of the sides of the special triangle to evaluate the ratio $\sin30°$ .

STEP: Draw a special triangle and use it to evaluate the expression

\sin 30 °

$\sin30°$
[−1 point ⇒ 0 / 1 points left]

The question asks us to evaluate the trigonometric expression

\begin{matrix} \sin 30 ° \end{matrix}

$\begin{array}{c}\sin30°\end{array}$

One way to find the answer would be to type this into a calculator. But the question specifically says no calculators!

Lucky for us, $30°$ is a special angle! Special angles are angles for which we can evaluate the trigonometric ratios exactly.

We can do this using the special triangles. The special triangle that contains the angle $30°$ is shown below. Since we want to find $\sin30°$ , we need to identify the opposite and the hypotenuse.

Taking the ratio of the opposite and the hypotenuse, we can see that

\begin{matrix} \sin 30 ° = \frac{1}{2} \end{matrix}

$\begin{array}{c}\sin30° = \dfrac{1}{2}\end{array}$

The correct answer is D: $\sin30° = \dfrac{1}{2}$ .

Submit your answer as:

ID is: 3521 Seed is: 8308

Special angles and triangles

The diagram below shows a right-angled triangle. The acute angles are both equal to $45°$ . Side $\overline{TV}$ has length $1$ and side $\overline{VS}$ has length $\sqrt{2}$ .

Answer the following questions about this triangle.

What is the length of the missing side $\overline{ST}$ ?

INSTRUCTION: Type sqrt( ) if you need to show a square root.

Answer: $\overline{ST} =$
expression
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The triangle given in the question is a special triangle. You need to remember the angles and side lengths of the special triangles. If you can't remember them, have a look at them in the Everything Maths textbook here.

STEP: Remember the values of the sides of the special triangle
[−1 point ⇒ 0 / 1 points left]

The question gives us a right-angled triangle with two acute angles and two sides labelled. We need to find the length of the third side.

Since this is a right-angled triangle, we could use the theorem of Pythagoras to find the missing length. But we don't need to, because this is a special triangle.

The special triangles have sides that you need to know for tests and exams. You can find an explanation about special triangles in the Everything Maths textbook here.

So the length of the missing side is $\overline{ST} = 1$ .

Submit your answer as:
Using the triangle from Question 1, evaluate the following trigonometric expression:

$\begin{array}{c}\tan 45°\end{array}$

INSTRUCTION: Do not use a calculator. Type sqrt( ) if you need to show a square root.

Answer: $\tan 45° =$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Now that you have the lengths of all the sides of the triangle, you can determine the trigonometric ratio $\tan 45°$ directly from the triangle.

STEP: Evaluate the trigonometric ratio from the triangle
[−1 point ⇒ 0 / 1 points left]

The question asks us to evaluate the trigonometric expression

$\begin{array}{c}\tan 45°\end{array}$

One way to do this would be to type this into a calculator. But the instructions say no calculators!

In Question 1, we determined the length of the missing side of the given special triangle. So we can evaluate $\tan 45°$ directly from the triangle by taking the ratio of the opposite and the adjacent for the $45°$ angle. These sides are shown in the diagram below.

Taking the ratio of the opposite and the adjacent, we can see that

$\begin{array}{c}\tan 45° = 1\end{array}$

NOTE: It does not matter which angle we use, because they are both $45°$ . If we use the other $45°$ angle, the answer would be the same.

So the correct answer is $\tan 45° = 1$ .

Submit your answer as:

ID is: 3521 Seed is: 7308

Special angles and triangles

The diagram below shows a right-angled triangle. The acute angles are both equal to $45°$ . Side $\overline{AB}$ has length $1$ and side $\overline{BC}$ has length $1$ .

Answer the following questions about this triangle.

What is the length of the missing side $\overline{CA}$ ?

INSTRUCTION: Type sqrt( ) if you need to show a square root.

Answer: $\overline{CA} =$
expression
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The triangle given in the question is a special triangle. You need to remember the angles and side lengths of the special triangles. If you can't remember them, have a look at them in the Everything Maths textbook here.

STEP: Remember the values of the sides of the special triangle
[−1 point ⇒ 0 / 1 points left]

The question gives us a right-angled triangle with two acute angles and two sides labelled. We need to find the length of the third side.

Since this is a right-angled triangle, we could use the theorem of Pythagoras to find the missing length. But we don't need to, because this is a special triangle.

The special triangles have sides that you need to know for tests and exams. You can find an explanation about special triangles in the Everything Maths textbook here.

So the length of the missing side is $\overline{CA} = \sqrt{2}$ .

Submit your answer as:
Using the triangle from Question 1, evaluate the following trigonometric expression:

$\begin{array}{c}\sin 45°\end{array}$

INSTRUCTION: Do not use a calculator. Type sqrt( ) if you need to show a square root.

Answer: $\sin 45° =$
expression
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Now that you have the lengths of all the sides of the triangle, you can determine the trigonometric ratio $\sin 45°$ directly from the triangle.

STEP: Evaluate the trigonometric ratio from the triangle
[−1 point ⇒ 0 / 1 points left]

The question asks us to evaluate the trigonometric expression

$\begin{array}{c}\sin 45°\end{array}$

One way to do this would be to type this into a calculator. But the instructions say no calculators!

In Question 1, we determined the length of the missing side of the given special triangle. So we can evaluate $\sin 45°$ directly from the triangle by taking the ratio of the opposite and the hypotenuse for the $45°$ angle. These sides are shown in the diagram below.

Taking the ratio of the opposite and the hypotenuse, we can see that

$\begin{array}{c}\sin 45° = \frac{ 1 }{ \sqrt{2} }\end{array}$

NOTE: It does not matter which angle we use, because they are both $45°$ . If we use the other $45°$ angle, the answer would be the same.

So the correct answer is $\sin 45° = \dfrac{ 1 }{ \sqrt{2} }$ .

Submit your answer as:

ID is: 3521 Seed is: 8288

Special angles and triangles

The diagram below shows a right-angled triangle. The acute angles are equal to $30°$ and $60°$ . Side $\overline{PQ}$ has length $1$ and side $\overline{RP}$ has length $2$ .

Answer the following questions about this triangle.

What is the length of the missing side $\overline{QR}$ ?

INSTRUCTION: Type sqrt( ) if you need to show a square root.

Answer: $\overline{QR} =$
expression
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The triangle given in the question is a special triangle. You need to remember the angles and side lengths of the special triangles. If you can't remember them, have a look at them in the Everything Maths textbook here.

STEP: Remember the values of the sides of the special triangle
[−1 point ⇒ 0 / 1 points left]

The question gives us a right-angled triangle with two acute angles and two sides labelled. We need to find the length of the third side.

Since this is a right-angled triangle, we could use the theorem of Pythagoras to find the missing length. But we don't need to, because this is a special triangle.

The special triangles have sides that you need to know for tests and exams. You can find an explanation about special triangles in the Everything Maths textbook here.

So the length of the missing side is $\overline{QR} = \sqrt{3}$ .

Submit your answer as:
Using the triangle from Question 1, evaluate the following trigonometric expression:

$\begin{array}{c}\tan 60°\end{array}$

INSTRUCTION: Do not use a calculator. Type sqrt( ) if you need to show a square root.

Answer: $\tan 60° =$
expression
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Now that you have the lengths of all the sides of the triangle, you can determine the trigonometric ratio $\tan 60°$ directly from the triangle.

STEP: Evaluate the trigonometric ratio from the triangle
[−1 point ⇒ 0 / 1 points left]

The question asks us to evaluate the trigonometric expression

$\begin{array}{c}\tan 60°\end{array}$

One way to do this would be to type this into a calculator. But the instructions say no calculators!

In Question 1, we determined the length of the missing side of the given special triangle. So we can evaluate $\tan 60°$ directly from the triangle by taking the ratio of the opposite and the adjacent for the $60°$ angle. These sides are shown in the diagram below.

Taking the ratio of the opposite and the adjacent, we can see that

$\begin{array}{c}\tan 60° = \sqrt{3}\end{array}$

So the correct answer is $\tan 60° = \sqrt{3}$ .

Submit your answer as:

3. Practical applications

4. Trigonometry on a unit circle

ID is: 3662 Seed is: 4525

Trigonometry on the Cartesian plane

Suppose $\cos \left(\theta\right) = \dfrac{5}{13}$ , and $0° \leq \theta \leq 180°$ . Answer the three questions below about this equation.

The angle $\theta$ points into one of the quadrants of the Cartesian plane. Which quadrant corresponds to $\theta$ ?

Answer: The angle $\theta$ points into Quadrant .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Compare the information given to the CAST diagram.

STEP: Compare the information given to the CAST diagram
[−1 point ⇒ 0 / 1 points left]

We need to figure out which quadrant the angle $\theta$ points to. We can answer this question using the CAST diagram because we already know that $\cos (\theta)$ is positive.

$\begin{array}{c} \cos (\theta) = \dfrac{5}{13} \end{array}$

The CAST diagram tells us that the cosine ratio is positive in both Quadrants I and IV. We also know that $\theta$ is in a specific interval: $0° \leq \theta \leq 180°$ .

The CAST diagram above shows that $\theta$ must point into Quadrant I. That is the only quadrant which has the correct sign (positive) and is also in the allowed interval.

NOTE: If we did not know that $0° \leq \theta \leq 180°$ we could not know which quadrant the angle is in!

The angle $\theta$ must point into Quadrant I.

Submit your answer as:
On the Cartesian plane, the trigonometric ratios are defined in terms of the coordinates $(x; y)$ and the radius $r$ . The equation $\cos (\theta) = \dfrac{5}{13}$ tells us about the values of $x$ and $r$ . For angle $\theta$ , what is the value of the $y$ -coordinate?

Answer: The value of the $y$ -coordinate is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by drawing a diagram to represent the angle $\theta$ . Remember from Question 1 you know the angle must be in Quadrant I.

STEP: Draw a diagram to represent the angle
[−1 point ⇒ 2 / 3 points left]

Start by drawing a triangle on the Cartesian plane to represent the angle $\theta$ . From Question 1, we know that $\theta$ must be in Quadrant I. We will put a point in that quadrant and label it $P$ . We will also draw a reference triangle for this point. This should be a right-angled triangle with the right-angle on the $x$ -axis.

NOTE: We do not know the angle or the exact position of the point: we only know that they must be in Quadrant I. So the diagram above is not precise: it is a sketch. Since the sketch is not precise, we cannot trust the appearance of the figure. Instead, we can only trust the labels.

STEP: Determine the values available from $\cos (\theta) = \dfrac{5}{13}$
[−1 point ⇒ 1 / 3 points left]

For angle $\theta$ , we know that the cosine ratio is equal to $\dfrac{5}{13}$ . Comparing this to the definition of the cosine ratio, we can read off the values of $x$ and $r$ :

$\begin{aligned} \cos (\theta) = \dfrac{5}{13} & = \dfrac{x}{r} \\ & \downarrow \\ \dfrac{x}{r} & = \dfrac{5}{13} \\ \text{This means:} & \\ x & = 5 \\ r & = 13 \end{aligned}$

Now we can label two sides of the triangle, and one of the coordinates of Point $P$ .

NOTE: The lengths of the triangle's sides must be positive because they are distances. But the coordinates of Point $P$ can be positive or negative, depending on which quadrant the point is in. It is crucial to be aware that these signs can be different.

STEP: Calculate the third side of the triangle
[−1 point ⇒ 0 / 3 points left]

Now we can calculate the value of $y$ for angle $\theta$ . We do this using the reference triangle: use the theorem of Pythagoras.

$\begin{aligned} a^{2} + b^{2} & = c^{2} \\ y^{2} + (5)^{2} & = (13)^{2} \\ y^{2} & = 169 - 25 \\ y^{2} & = 144 \\ y & = \pm 12 \end{aligned}$

The length of a side of a triangle is always positive. And since Point $P$ is in Quadrant I, the $y$ -coordinate is going to be positive as well.

The value of the $y$ -coordinate is $\text{12}$ .

Submit your answer as:
Use the result from Question 2 to determine the value of $\tan (\theta)$ .

Answer: $\tan (\theta) =$ .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to combine the result of Question 2 with the definition of $\tan (\theta)$ .

STEP: Use the definition of the tangent ratio on the Cartesian plane
[−2 points ⇒ 0 / 2 points left]

The definition of the tangent ratio on the Cartesian plane is:

$\begin{array}{c} \tan (\theta) = \dfrac{y}{x} \end{array}$

We know both $y$ and $x$ for angle $\theta$ from the information we got in Question 2. All we need to do is substitute in the values for the point corresponding to $\theta$ , which is Point $P$ .

$\begin{aligned} \tan (\theta) & = \dfrac{y}{x} \\ & = \frac{12}{5} \end{aligned}$

TIP: Compare the sign of your answer to the CAST diagram. Point $P$ is in Quadrant I. And the CAST diagram tells us that the tangent ratio in Quadrant I is always positive. Make sure the sign of your answer agrees with the CAST diagram.

The value of $\tan (\theta)$ is $\dfrac{12}{5}$ .

Submit your answer as:

ID is: 3662 Seed is: 4149

Trigonometry on the Cartesian plane

Suppose $\cos \left(\theta\right) = \dfrac{15}{17}$ , and $180° \leq \theta \leq 360°$ . Answer the three questions below about this equation.

The angle $\theta$ points into one of the quadrants of the Cartesian plane. Which quadrant corresponds to $\theta$ ?

Answer: The angle $\theta$ points into Quadrant .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Compare the information given to the CAST diagram.

STEP: Compare the information given to the CAST diagram
[−1 point ⇒ 0 / 1 points left]

We need to figure out which quadrant the angle $\theta$ points to. We can answer this question using the CAST diagram because we already know that $\cos (\theta)$ is positive.

$\begin{array}{c} \cos (\theta) = \dfrac{15}{17} \end{array}$

The CAST diagram tells us that the cosine ratio is positive in both Quadrants I and IV. We also know that $\theta$ is in a specific interval: $180° \leq \theta \leq 360°$ .

The CAST diagram above shows that $\theta$ must point into Quadrant IV. That is the only quadrant which has the correct sign (positive) and is also in the allowed interval.

NOTE: If we did not know that $180° \leq \theta \leq 360°$ we could not know which quadrant the angle is in!

The angle $\theta$ must point into Quadrant IV.

Submit your answer as:
On the Cartesian plane, the trigonometric ratios are defined in terms of the coordinates $(x; y)$ and the radius $r$ . The equation $\cos (\theta) = \dfrac{15}{17}$ tells us about the values of $x$ and $r$ . For angle $\theta$ , what is the value of the $y$ -coordinate?

Answer: The value of the $y$ -coordinate is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by drawing a diagram to represent the angle $\theta$ . Remember from Question 1 you know the angle must be in Quadrant IV.

STEP: Draw a diagram to represent the angle
[−1 point ⇒ 2 / 3 points left]

Start by drawing a triangle on the Cartesian plane to represent the angle $\theta$ . From Question 1, we know that $\theta$ must be in Quadrant IV. We will put a point in that quadrant and label it $P$ . We will also draw a reference triangle for this point. This should be a right-angled triangle with the right-angle on the $x$ -axis.

NOTE: We do not know the angle or the exact position of the point: we only know that they must be in Quadrant IV. So the diagram above is not precise: it is a sketch. Since the sketch is not precise, we cannot trust the appearance of the figure. Instead, we can only trust the labels.

STEP: Determine the values available from $\cos (\theta) = \dfrac{15}{17}$
[−1 point ⇒ 1 / 3 points left]

For angle $\theta$ , we know that the cosine ratio is equal to $\dfrac{15}{17}$ . Comparing this to the definition of the cosine ratio, we can read off the values of $x$ and $r$ :

$\begin{aligned} \cos (\theta) = \dfrac{15}{17} & = \dfrac{x}{r} \\ & \downarrow \\ \dfrac{x}{r} & = \dfrac{15}{17} \\ \text{This means:} & \\ x & = 15 \\ r & = 17 \end{aligned}$

The hypotenuse $(r)$ is always positive, so the $x$ -coordinate must be negative. This makes sense because we know the $x$ -coordinate is always negative in Quadrant IV.

Now we can label two sides of the triangle, and one of the coordinates of Point $P$ .

NOTE: The lengths of the triangle's sides must be positive because they are distances. But the coordinates of Point $P$ can be positive or negative, depending on which quadrant the point is in. It is crucial to be aware that these signs can be different.

STEP: Calculate the third side of the triangle
[−1 point ⇒ 0 / 3 points left]

Now we can calculate the value of $y$ for angle $\theta$ . We do this using the reference triangle: use the theorem of Pythagoras.

$\begin{aligned} a^{2} + b^{2} & = c^{2} \\ y^{2} + (15)^{2} & = (17)^{2} \\ y^{2} & = 289 - 225 \\ y^{2} & = 64 \\ y & = \pm 8 \end{aligned}$

The length of a side of a triangle is always positive. But since Point $P$ is in Quadrant IV, the $y$ -coordinate is going to be negative.

The value of the $y$ -coordinate is $-\text{8}$ .

Submit your answer as:
Use the result from Question 2 to determine the value of $\sin (\theta)$ .

Answer: $\sin (\theta) =$ .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to combine the result of Question 2 with the definition of $\sin (\theta)$ .

STEP: Use the definition of the sine ratio on the Cartesian plane
[−2 points ⇒ 0 / 2 points left]

The definition of the sine ratio on the Cartesian plane is:

$\begin{array}{c} \sin (\theta) = \dfrac{y}{r} \end{array}$

We know both $y$ and $r$ for angle $\theta$ from the information we got in Question 2. All we need to do is substitute in the values for the point corresponding to $\theta$ , which is Point $P$ .

$\begin{aligned} \sin (\theta) & = \dfrac{y}{r} \\ & = - \frac{8}{17} \end{aligned}$

TIP: Compare the sign of your answer to the CAST diagram. Point $P$ is in Quadrant IV. And the CAST diagram tells us that the sine ratio in Quadrant IV is always negative. Make sure the sign of your answer agrees with the CAST diagram.

The value of $\sin (\theta)$ is $- \dfrac{8}{17}$ .

Submit your answer as:

ID is: 3662 Seed is: 9046

Trigonometry on the Cartesian plane

Suppose $\sin \left(\theta\right) = \dfrac{5}{13}$ , and $90° \leq \theta \leq 270°$ . Answer the three questions below about this equation.

The angle $\theta$ points into one of the quadrants of the Cartesian plane. Which quadrant corresponds to $\theta$ ?

Answer: The angle $\theta$ points into Quadrant .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Compare the information given to the CAST diagram.

STEP: Compare the information given to the CAST diagram
[−1 point ⇒ 0 / 1 points left]

We need to figure out which quadrant the angle $\theta$ points to. We can answer this question using the CAST diagram because we already know that $\sin (\theta)$ is positive.

$\begin{array}{c} \sin (\theta) = \dfrac{5}{13} \end{array}$

The CAST diagram tells us that the sine ratio is positive in both Quadrants I and II. We also know that $\theta$ is in a specific interval: $90° \leq \theta \leq 270°$ .

The CAST diagram above shows that $\theta$ must point into Quadrant II. That is the only quadrant which has the correct sign (positive) and is also in the allowed interval.

NOTE: If we did not know that $90° \leq \theta \leq 270°$ we could not know which quadrant the angle is in!

The angle $\theta$ must point into Quadrant II.

Submit your answer as:
On the Cartesian plane, the trigonometric ratios are defined in terms of the coordinates $(x; y)$ and the radius $r$ . The equation $\sin (\theta) = \dfrac{5}{13}$ tells us about the values of $y$ and $r$ . For angle $\theta$ , what is the value of the $x$ -coordinate?

Answer: The value of the $x$ -coordinate is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by drawing a diagram to represent the angle $\theta$ . Remember from Question 1 you know the angle must be in Quadrant II.

STEP: Draw a diagram to represent the angle
[−1 point ⇒ 2 / 3 points left]

Start by drawing a triangle on the Cartesian plane to represent the angle $\theta$ . From Question 1, we know that $\theta$ must be in Quadrant II. We will put a point in that quadrant and label it $P$ . We will also draw a reference triangle for this point. This should be a right-angled triangle with the right-angle on the $x$ -axis.

NOTE: We do not know the angle or the exact position of the point: we only know that they must be in Quadrant II. So the diagram above is not precise: it is a sketch. Since the sketch is not precise, we cannot trust the appearance of the figure. Instead, we can only trust the labels.

STEP: Determine the values available from $\sin (\theta) = \dfrac{5}{13}$
[−1 point ⇒ 1 / 3 points left]

For angle $\theta$ , we know that the sine ratio is equal to $\dfrac{5}{13}$ . Comparing this to the definition of the sine ratio, we can read off the values of $y$ and $r$ :

$\begin{aligned} \sin (\theta) = \dfrac{5}{13} & = \dfrac{y}{r} \\ & \downarrow \\ \dfrac{y}{r} & = \dfrac{5}{13} \\ \text{This means:} & \\ y & = 5 \\ r & = 13 \end{aligned}$

The hypotenuse $(r)$ is always positive, so the $y$ -coordinate must be negative. This makes sense because we know the $y$ -coordinate is always negative in Quadrant II.

Now we can label two sides of the triangle, and one of the coordinates of Point $P$ .

NOTE: The lengths of the triangle's sides must be positive because they are distances. But the coordinates of Point $P$ can be positive or negative, depending on which quadrant the point is in. It is crucial to be aware that these signs can be different.

STEP: Calculate the third side of the triangle
[−1 point ⇒ 0 / 3 points left]

Now we can calculate the value of $x$ for angle $\theta$ . We do this using the reference triangle: use the theorem of Pythagoras.

$\begin{aligned} a^{2} + b^{2} & = c^{2} \\ x^{2} + (5)^{2} & = (13)^{2} \\ x^{2} & = 169 - 25 \\ x^{2} & = 144 \\ x & = \pm 12 \end{aligned}$

The length of a side of a triangle is always positive. But since Point $P$ is in Quadrant II, the $x$ -coordinate is going to be negative.

The value of the $x$ -coordinate is $-\text{12}$ .

Submit your answer as:
Use the result from Question 2 to determine the value of $\cos (\theta)$ .

Answer: $\cos (\theta) =$ .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to combine the result of Question 2 with the definition of $\cos (\theta)$ .

STEP: Use the definition of the cosine ratio on the Cartesian plane
[−2 points ⇒ 0 / 2 points left]

The definition of the cosine ratio on the Cartesian plane is:

$\begin{array}{c} \cos (\theta) = \dfrac{x}{r} \end{array}$

We know both $x$ and $r$ for angle $\theta$ from the information we got in Question 2. All we need to do is substitute in the values for the point corresponding to $\theta$ , which is Point $P$ .

$\begin{aligned} \cos (\theta) & = \dfrac{x}{r} \\ & = - \frac{12}{13} \end{aligned}$

TIP: Compare the sign of your answer to the CAST diagram. Point $P$ is in Quadrant II. And the CAST diagram tells us that the cosine ratio in Quadrant II is always negative. Make sure the sign of your answer agrees with the CAST diagram.

The value of $\cos (\theta)$ is $- \dfrac{12}{13}$ .

Submit your answer as:

ID is: 3573 Seed is: 8673

The trigonometric ratios on the Cartesian plane

On the Cartesian plane the trigonometric ratios are defined in terms of the coordinates $x$ and $y$ , and the radius $r$ . Complete the following equation to make it true (identify the value which belongs in place of the ?):

\begin{matrix} \sin θ = \frac{y}{?} \end{matrix}

$\begin{array}{c} \sin \theta = \dfrac{ y } { ? } \end{array}$

Answer: The missing quantity is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to use the definition of sine on the Cartesian plane.

STEP: Use the definition of sine on the Cartesian plane
[−1 point ⇒ 0 / 1 points left]

We use the opposite, adjacent, and hypotenuse sides of a right-angled triangle to define the trigonometric ratios. But we can translate these definitions onto the Cartesian plane using a right-angled triangle on the Cartesian plane.

The diagram below shows a point on the circle at $(x; y)$ . The radius of the circle is $r$ . The point makes an angle $\theta$ with the positive $x$ -axis. The sides of the triangle have lengths $x$ , $y$ , and $r$ .

Here are three key relationships in the triangle above:

The hypotenuse of the triangle is the radius of the circle.
The side of the triangle opposite to $\theta$ has a length $y$ .
The side of the triangle adjacent to $\theta$ has a length $x$ .

Using the definitions of the trigonometric ratios in terms of the opposite, adjacent, and hypotenuse sides of the triangle, we can write:

\begin{aligned} \sin θ = & \frac{opposite}{hypotenuse} & \to \frac{y}{r} \\ \cos θ = & \frac{adjacent}{hypotenuse} & \to \frac{x}{r} \\ \tan θ = & \frac{opposite}{adjacent} & \to \frac{y}{x} \end{aligned}

$\begin{alignedat}{2} \sin \theta = & \dfrac{\text{opposite}}{\text{hypotenuse}} \; & \rightarrow \; \dfrac{y}{r} \\[0.2cm] \cos \theta = & \dfrac{\text{adjacent}}{\text{hypotenuse}} \; & \rightarrow \; \dfrac{x}{r} \\[0.2cm] \tan \theta = & \dfrac{\text{opposite}}{\text{adjacent}} \; & \rightarrow \; \dfrac{y}{x} \end{alignedat}$

By comparing these to the equation in the question we can find the missing value.

The correct definition is $\sin \theta = \dfrac{ y } { r }$ so the missing quantity is $r$ .

Submit your answer as:

ID is: 3573 Seed is: 9366

The trigonometric ratios on the Cartesian plane

\begin{matrix} \cos θ = \frac{x}{?} \end{matrix}

$\begin{array}{c} \cos \theta = \dfrac{ x } { ? } \end{array}$

Answer: The missing quantity is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to use the definition of cosine on the Cartesian plane.

STEP: Use the definition of cosine on the Cartesian plane
[−1 point ⇒ 0 / 1 points left]

Here are three key relationships in the triangle above:

The hypotenuse of the triangle is the radius of the circle.
The side of the triangle opposite to $\theta$ has a length $y$ .
The side of the triangle adjacent to $\theta$ has a length $x$ .

Using the definitions of the trigonometric ratios in terms of the opposite, adjacent, and hypotenuse sides of the triangle, we can write:

\begin{aligned} \sin θ = & \frac{opposite}{hypotenuse} & \to \frac{y}{r} \\ \cos θ = & \frac{adjacent}{hypotenuse} & \to \frac{x}{r} \\ \tan θ = & \frac{opposite}{adjacent} & \to \frac{y}{x} \end{aligned}

By comparing these to the equation in the question we can find the missing value.

The correct definition is $\cos \theta = \dfrac{ x } { r }$ so the missing quantity is $r$ .

Submit your answer as:

ID is: 3573 Seed is: 9458

The trigonometric ratios on the Cartesian plane

\begin{matrix} \cos θ = \frac{?}{r} \end{matrix}

$\begin{array}{c} \cos \theta = \dfrac{ ? } { r } \end{array}$

Answer: The missing quantity is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to use the definition of cosine on the Cartesian plane.

STEP: Use the definition of cosine on the Cartesian plane
[−1 point ⇒ 0 / 1 points left]

Here are three key relationships in the triangle above:

The hypotenuse of the triangle is the radius of the circle.
The side of the triangle opposite to $\theta$ has a length $y$ .
The side of the triangle adjacent to $\theta$ has a length $x$ .

Using the definitions of the trigonometric ratios in terms of the opposite, adjacent, and hypotenuse sides of the triangle, we can write:

\begin{aligned} \sin θ = & \frac{opposite}{hypotenuse} & \to \frac{y}{r} \\ \cos θ = & \frac{adjacent}{hypotenuse} & \to \frac{x}{r} \\ \tan θ = & \frac{opposite}{adjacent} & \to \frac{y}{x} \end{aligned}

By comparing these to the equation in the question we can find the missing value.

The correct definition is $\cos \theta = \dfrac{ x } { r }$ so the missing quantity is $x$ .

Submit your answer as:

ID is: 3597 Seed is: 6768

Reading ratios from the Cartesian plane

The diagram below shows Point $P$ at $(-6; -8)$ . A triangle is also shown. The lengths of the triangle's sides are $6$ , $8$ , and $10$ , as labelled. This triangle contains an angle which is labelled $\theta$ . The angle $\alpha$ is also labelled between the positive $x$ -axis around to the hypotenuse of the triangle. Answer the two questions which follow.

Based on the diagram, what is the value of $\cos \theta$ ?

INSTRUCTION: Give your answer in the form of a fraction.

Answer: $\cos \theta =$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start with the fact that the cosine ratio is always the adjacent over the hypotenuse. You can find those values on the triangle in the figure.

STEP: Read the answer from the triangle in the figure
[−1 point ⇒ 0 / 1 points left]

We need to find the value of $\cos \theta$ . The cosine ratio is always the adjacent over the hypotenuse. So we can find the answer from the labels on the triangle.

Now write down the ratio:

$\begin{aligned} \cos \theta & = \frac{6}{10} \\[0.1cm] & = \dfrac{3}{5} \end{aligned}$

NOTE: For the angle $\theta$ , we needed only the labels in the triangle. This will be different in Question 2.

The value of $\cos \theta$ is $\dfrac{3}{5}$ .

Submit your answer as:
What is the value of $\cos \alpha$ ?

INSTRUCTION: Give your answer in the form of a fraction.
Answer: $\cos \alpha =$ .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the definition of the cosine ratio on the Cartesian plane.

STEP: Write down the definition of the cosine ratio on the Cartesian plane
[−1 point ⇒ 1 / 2 points left]

In Question 1 we were working with the angle $\theta$ . But now we are working with the angle $\alpha$ . So we need to use the definition of cosine on the Cartesian plane. This is for two reasons:
- Unlike $\theta$ , the angle $\alpha$ is not inside of the triangle. So we cannot read off the answer from the triangle like we did in Question 1.
- The trigonometric ratios on the Cartesian plane are defined with the angle starting on the positive $x$ -axis, and that is where the angle $\alpha$ starts.
The definition for the cosine ratio is:

$\begin{array}{c} \cos \alpha = \dfrac{x}{r} \end{array}$

STEP: Substitute the values for $x$ and $r$
[−1 point ⇒ 0 / 2 points left]

To continue, we must substitute in the correct values. Note that these are not the same values we used in Question 1. The definition here refers to the $x$ -coordinate of Point $P$ , which is $-\text{6}$ .

$\begin{alignedat}{1} \cos \alpha &= \dfrac{x}{r} \\ &\downarrow \\ \text{For Point } P, \; &x = -6, r = 10 \\ &\downarrow \\ \cos \alpha &= \dfrac{-6}{10} \\ &= - \dfrac{3}{5} \end{alignedat}$

It is now clear that $\cos \alpha$ is not the same as $\cos \theta$ . But the numbers in each ratio are the same: only the signs are different. This is because the angles are closely related: they are both connected to Point $P$ . $\theta$ is called a reference angle for $\alpha$ . And the triangle in the question is called a reference triangle. The reference triangle is useful because it allows us to find the value - but not the sign - for any ratio for the angle $\alpha$ .

NOTE: Sometimes the signs of the ratios for the reference angle and the full angle are the same, and sometimes they are opposites. But the numbers are always the same. In other words, the reference angle and the full angle will always lead to the same numbers, but the signs might be different. The sign for $\cos \alpha$ depends on which quadrant the reference triangle is in (the size of $\alpha$ ). You can answer this question using the first answer, which is always positive, and the CAST diagram. The CAST diagram tells us that the cosine ratio is negative in the third quadrant, which tells us that $\cos \alpha$ must be negative.

The value of $\cos \alpha$ is $- \dfrac{3}{5}$ .

Submit your answer as:

ID is: 3597 Seed is: 2446

Reading ratios from the Cartesian plane

The diagram below shows Point $P$ at $(6; -8)$ . A triangle is also shown. The lengths of the triangle's sides are $6$ , $8$ , and $10$ , as labelled. This triangle contains an angle which is labelled $\theta$ . The angle $\alpha$ is also labelled between the positive $x$ -axis around to the hypotenuse of the triangle. Answer the two questions which follow.

Based on the diagram, what is the value of $\cos \theta$ ?

INSTRUCTION: Give your answer in the form of a fraction.

Answer: $\cos \theta =$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start with the fact that the cosine ratio is always the adjacent over the hypotenuse. You can find those values on the triangle in the figure.

STEP: Read the answer from the triangle in the figure
[−1 point ⇒ 0 / 1 points left]

We need to find the value of $\cos \theta$ . The cosine ratio is always the adjacent over the hypotenuse. So we can find the answer from the labels on the triangle.

Now write down the ratio:

$\begin{aligned} \cos \theta & = \frac{6}{10} \\[0.1cm] & = \dfrac{3}{5} \end{aligned}$

NOTE: For the angle $\theta$ , we needed only the labels in the triangle. This will be different in Question 2.

The value of $\cos \theta$ is $\dfrac{3}{5}$ .

Submit your answer as:
What is the value of $\cos \alpha$ ?

INSTRUCTION: Give your answer in the form of a fraction.
Answer: $\cos \alpha =$ .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the definition of the cosine ratio on the Cartesian plane.

STEP: Write down the definition of the cosine ratio on the Cartesian plane
[−1 point ⇒ 1 / 2 points left]

In Question 1 we were working with the angle $\theta$ . But now we are working with the angle $\alpha$ . So we need to use the definition of cosine on the Cartesian plane. This is for two reasons:
- Unlike $\theta$ , the angle $\alpha$ is not inside of the triangle. So we cannot read off the answer from the triangle like we did in Question 1.
- The trigonometric ratios on the Cartesian plane are defined with the angle starting on the positive $x$ -axis, and that is where the angle $\alpha$ starts.
The definition for the cosine ratio is:

$\begin{array}{c} \cos \alpha = \dfrac{x}{r} \end{array}$

STEP: Substitute the values for $x$ and $r$
[−1 point ⇒ 0 / 2 points left]

To continue, we must substitute in the correct values. Note that these are not the same values we used in Question 1. The definition here refers to the $x$ -coordinate of Point $P$ , which is $\text{6}$ .

$\begin{alignedat}{1} \cos \alpha &= \dfrac{x}{r} \\ &\downarrow \\ \text{For Point } P, \; &x = 6, r = 10 \\ &\downarrow \\ \cos \alpha &= \dfrac{6}{10} \\ &= \dfrac{3}{5} \end{alignedat}$

It is now clear that $\cos \alpha$ is the same as $\cos \theta$ . In this calculation both parts of the ratio were negative, but the ratio ends up positive. That means we get the same answer as in Question 1. This is because the angles are closely related: they are both connected to Point $P$ . $\theta$ is called a reference angle for $\alpha$ . And the triangle in the question is called a reference triangle. The reference triangle is useful because it allows us to find the value - but not the sign - for any ratio for the angle $\alpha$ .

NOTE: Sometimes the signs of the ratios for the reference angle and the full angle are the same, and sometimes they are opposites. But the numbers are always the same. In other words, the reference angle and the full angle will always lead to the same numbers, but the signs might be different. The sign for $\cos \alpha$ depends on which quadrant the reference triangle is in (the size of $\alpha$ ). You can answer this question using the first answer, which is always positive, and the CAST diagram. The CAST diagram tells us that the cosine ratio is positive in the fourth quadrant, which tells us that $\cos \alpha$ must be positive.

The value of $\cos \alpha$ is $\dfrac{3}{5}$ .

Submit your answer as:

ID is: 3597 Seed is: 9061

Reading ratios from the Cartesian plane

The diagram below shows Point $P$ at $(8; -6)$ . A triangle is also shown. The lengths of the triangle's sides are $8$ , $6$ , and $10$ , as labelled. This triangle contains an angle which is labelled $\theta$ . The angle $\alpha$ is also labelled between the positive $x$ -axis around to the hypotenuse of the triangle. Answer the two questions which follow.

Based on the diagram, what is the value of $\cos \theta$ ?

INSTRUCTION: Give your answer in the form of a fraction.

Answer: $\cos \theta =$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start with the fact that the cosine ratio is always the adjacent over the hypotenuse. You can find those values on the triangle in the figure.

STEP: Read the answer from the triangle in the figure
[−1 point ⇒ 0 / 1 points left]

We need to find the value of $\cos \theta$ . The cosine ratio is always the adjacent over the hypotenuse. So we can find the answer from the labels on the triangle.

Now write down the ratio:

$\begin{aligned} \cos \theta & = \frac{8}{10} \\[0.1cm] & = \dfrac{4}{5} \end{aligned}$

NOTE: For the angle $\theta$ , we needed only the labels in the triangle. This will be different in Question 2.

The value of $\cos \theta$ is $\dfrac{4}{5}$ .

Submit your answer as:
What is the value of $\cos \alpha$ ?

INSTRUCTION: Give your answer in the form of a fraction.
Answer: $\cos \alpha =$ .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the definition of the cosine ratio on the Cartesian plane.

STEP: Write down the definition of the cosine ratio on the Cartesian plane
[−1 point ⇒ 1 / 2 points left]

In Question 1 we were working with the angle $\theta$ . But now we are working with the angle $\alpha$ . So we need to use the definition of cosine on the Cartesian plane. This is for two reasons:
- Unlike $\theta$ , the angle $\alpha$ is not inside of the triangle. So we cannot read off the answer from the triangle like we did in Question 1.
- The trigonometric ratios on the Cartesian plane are defined with the angle starting on the positive $x$ -axis, and that is where the angle $\alpha$ starts.
The definition for the cosine ratio is:

$\begin{array}{c} \cos \alpha = \dfrac{x}{r} \end{array}$

STEP: Substitute the values for $x$ and $r$
[−1 point ⇒ 0 / 2 points left]

To continue, we must substitute in the correct values. Note that these are not the same values we used in Question 1. The definition here refers to the $x$ -coordinate of Point $P$ , which is $\text{8}$ .

$\begin{alignedat}{1} \cos \alpha &= \dfrac{x}{r} \\ &\downarrow \\ \text{For Point } P, \; &x = 8, r = 10 \\ &\downarrow \\ \cos \alpha &= \dfrac{8}{10} \\ &= \dfrac{4}{5} \end{alignedat}$

It is now clear that $\cos \alpha$ is the same as $\cos \theta$ . In this calculation both parts of the ratio were negative, but the ratio ends up positive. That means we get the same answer as in Question 1. This is because the angles are closely related: they are both connected to Point $P$ . $\theta$ is called a reference angle for $\alpha$ . And the triangle in the question is called a reference triangle. The reference triangle is useful because it allows us to find the value - but not the sign - for any ratio for the angle $\alpha$ .

NOTE: Sometimes the signs of the ratios for the reference angle and the full angle are the same, and sometimes they are opposites. But the numbers are always the same. In other words, the reference angle and the full angle will always lead to the same numbers, but the signs might be different. The sign for $\cos \alpha$ depends on which quadrant the reference triangle is in (the size of $\alpha$ ). You can answer this question using the first answer, which is always positive, and the CAST diagram. The CAST diagram tells us that the cosine ratio is positive in the fourth quadrant, which tells us that $\cos \alpha$ must be positive.

The value of $\cos \alpha$ is $\dfrac{4}{5}$ .

Submit your answer as:

ID is: 1509 Seed is: 4678

Simplifying trigonometric expressions with reduction formulas

Without the use of a calculator, evaluate the trigonometric function shown.

\begin{matrix} \tan (405 °) \end{matrix}

$\begin{array}{c}\tan\left ( 405°\right )\end{array}$

INSTRUCTIONS:

Give your answer in surd form if necessary. You can type a surd like this: sqrt(10).
Do not use a calculator. While you can type the question into your calculator to get the answer, you will only get full marks in tests and exams if you show the necessary steps.

Answer:

\tan (405 °) =

$\tan\left ( 405°\right ) =$

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need to rewrite the angle into two terms. One of these parts should be one of $90°, 180°,$ or $360°$ . The other part should be one of the special angles, $30°, 45°,$ or $60°$ .

STEP: Rewrite the angle using one of the special angles
[−1 point ⇒ 2 / 3 points left]

The words 'without the use of a calculator' mean two important things:

We must show the working step by step.
We can expect to use special angles to solve the problem. The special angles are $30°, 45°,$ and $60°$ .

The first step is to break the angle in the question into two terms so that we can use one of the reduction formulas. We want one of those pieces to be a special angle because that is the only way we can evaluate the expression without a calculator. In this case, the original angle is $\text{405} \text{°}$ . We can split that into the values $\text{360} \text{°}$ and $\text{45} \text{°}$ using addition.

\begin{aligned} \tan (405 °) & = \tan (360 ° + 45 °) \end{aligned}

$\begin{aligned} \tan\left ( 405°\right ) & = \tan\left ( 360° +45°\right ) \end{aligned}$

STEP: Rewrite the expression using a reduction equation
[−1 point ⇒ 1 / 3 points left]

Now we can apply a reduction formula to simplify the angle in the expression. Specifically, we can remove the $\text{360} \text{°}$ part of the angle using the formula:

\begin{aligned} \tan (360 ° + θ) & = \tan θ \end{aligned}

$\begin{aligned} \tan\left(360° + \theta\right) & = \tan \theta \end{aligned}$

TIP: If you cannot remember the correct sign for the answer in the reduction formula, draw a quick picture of the CAST diagram.

Now we need to apply the formula above to the expression $\tan\left ( 360° +45°\right )$ .

\begin{aligned} \tan (360 ° + 45 °) & = \tan (45 °) \end{aligned}

$\begin{aligned} \tan\left ( 360° +45°\right ) & = \tan\left ( 45°\right ) \end{aligned}$

STEP: Evaluate the remaining expression
[−1 point ⇒ 0 / 3 points left]

Once we get the expression down to a special angle, we are expected to know the answer. That is why we can do the calculation without a calculator.

\begin{aligned} = \tan (45 °) \\ = 1 \end{aligned}

$\begin{aligned} & = \tan\left ( 45°\right ) \\ & = 1 \end{aligned}$

The final answer is $\tan\left ( 405°\right ) = 1$ .

Submit your answer as:

ID is: 1509 Seed is: 6567

Simplifying trigonometric expressions with reduction formulas

Determine the value of the following without using a calculator.

\begin{matrix} \sin (405 °) \end{matrix}

$\begin{array}{c}\sin\left ( 405°\right )\end{array}$

INSTRUCTIONS:

Give your answer in surd form if necessary. You can type a surd like this: sqrt(10).
Do not use a calculator. While you can type the question into your calculator to get the answer, you will only get full marks in tests and exams if you show the necessary steps.

Answer:

\sin (405 °) =

$\sin\left ( 405°\right ) =$

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need to rewrite the angle into two terms. One of these parts should be one of $90°, 180°,$ or $360°$ . The other part should be one of the special angles, $30°, 45°,$ or $60°$ .

STEP: Rewrite the angle using one of the special angles
[−1 point ⇒ 2 / 3 points left]

The words 'without the use of a calculator' mean two important things:

We must show the working step by step.
We can expect to use special angles to solve the problem. The special angles are $30°, 45°,$ and $60°$ .

\begin{aligned} \sin (405 °) & = \sin (360 ° + 45 °) \end{aligned}

$\begin{aligned} \sin\left ( 405°\right ) & = \sin\left ( 360° +45°\right ) \end{aligned}$

STEP: Rewrite the expression using a reduction equation
[−1 point ⇒ 1 / 3 points left]

Now we can apply a reduction formula to simplify the angle in the expression. Specifically, we can remove the $\text{360} \text{°}$ part of the angle using the formula:

\begin{aligned} \sin (360 ° + θ) & = \sin θ \end{aligned}

$\begin{aligned} \sin\left(360° + \theta\right) & = \sin \theta \end{aligned}$

TIP: If you cannot remember the correct sign for the answer in the reduction formula, draw a quick picture of the CAST diagram.

Now we need to apply the formula above to the expression $\sin\left ( 360° +45°\right )$ .

\begin{aligned} \sin (360 ° + 45 °) & = \sin (45 °) \end{aligned}

$\begin{aligned} \sin\left ( 360° +45°\right ) & = \sin\left ( 45°\right ) \end{aligned}$

STEP: Evaluate the remaining expression
[−1 point ⇒ 0 / 3 points left]

Once we get the expression down to a special angle, we are expected to know the answer. That is why we can do the calculation without a calculator.

\begin{aligned} = \sin (45 °) \\ = \frac{\sqrt{2}}{2} \end{aligned}

$\begin{aligned} & = \sin\left ( 45°\right ) \\ & = \dfrac{\sqrt{2}}{2} \end{aligned}$

The final answer is $\sin\left ( 405°\right ) = \dfrac{\sqrt{2}}{2}$ .

Submit your answer as:

ID is: 1509 Seed is: 2683

Simplifying trigonometric expressions with reduction formulas

Without the use of a calculator, evaluate the trigonometric function shown.

\begin{matrix} - \tan (120 °) \end{matrix}

$\begin{array}{c}- \tan\left ( 120°\right )\end{array}$

INSTRUCTIONS:

Give your answer in surd form if necessary. You can type a surd like this: sqrt(10).
Do not use a calculator. While you can type the question into your calculator to get the answer, you will only get full marks in tests and exams if you show the necessary steps.

Answer:

- \tan (120 °) =

$- \tan\left ( 120°\right ) =$

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need to rewrite the angle into two terms. One of these parts should be one of $90°, 180°,$ or $360°$ . The other part should be one of the special angles, $30°, 45°,$ or $60°$ .

STEP: Rewrite the angle using one of the special angles
[−1 point ⇒ 2 / 3 points left]

The words 'without the use of a calculator' mean two important things:

We must show the working step by step.
We can expect to use special angles to solve the problem. The special angles are $30°, 45°,$ and $60°$ .

The first step is to break the angle in the question into two terms so that we can use one of the reduction formulas. We want one of those pieces to be a special angle because that is the only way we can evaluate the expression without a calculator. In this case, the original angle is $\text{120} \text{°}$ . We can split that into the values $\text{180} \text{°}$ and $\text{60} \text{°}$ using subtraction.

\begin{aligned} - \tan (120 °) & = - \tan (180 ° - 60 °) \end{aligned}

$\begin{aligned} - \tan\left ( 120°\right ) & = - \tan\left ( 180° -60°\right ) \end{aligned}$

STEP: Rewrite the expression using a reduction equation
[−1 point ⇒ 1 / 3 points left]

Now we can apply a reduction formula to simplify the angle in the expression. Specifically, we can remove the $\text{180} \text{°}$ part of the angle using the formula:

\begin{aligned} \tan (180 ° - θ) & = - \tan θ \end{aligned}

$\begin{aligned} \tan\left(180° - \theta\right) & = -\tan \theta \end{aligned}$

TIP: If you cannot remember the correct sign for the answer in the reduction formula, draw a quick picture of the CAST diagram.

Now we need to apply the formula above to the expression $- \tan\left ( 180° -60°\right )$ . Note that there is a negative at the front of the expression. But the reduction formula above shows another negative will come into our solution. So we will find two negatives in the first step below.

\begin{aligned} - \tan (180 ° - 60 °) & = - (- \tan (60 °)) \\ = \tan (60 °) \end{aligned}

$\begin{aligned} - \tan\left ( 180° -60°\right ) & = - \left( - \tan\left ( 60°\right ) \right) \\ & = \tan\left ( 60°\right ) \end{aligned}$

STEP: Evaluate the remaining expression
[−1 point ⇒ 0 / 3 points left]

Once we get the expression down to a special angle, we are expected to know the answer. That is why we can do the calculation without a calculator.

\begin{aligned} = \tan (60 °) \\ = \sqrt{3} \end{aligned}

$\begin{aligned} & = \tan\left ( 60°\right ) \\ & = \sqrt{3} \end{aligned}$

The final answer is $- \tan\left ( 120°\right ) = \sqrt{3}$ .

Submit your answer as:

ID is: 3593 Seed is: 5484

Special angles: $0°$ and $90°$

On the Cartesian plane, the cosine ratio is defined as $\dfrac{x}{r}$ . Based on this definition or otherwise, determine the value of $\cos 90°$ .
INSTRUCTION:
- You may use a calculator.
- If the answer is undefined, type undefined.
Answer: The value of $\cos 90°$ is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you are not sure of the answer, use a calculator. That's right, we said it: use a calculator!

STEP: Evaluate the answer
[−1 point ⇒ 0 / 1 points left]

We need to determine the value of $\cos 90°$ . Here are two ways you might find the answer:
- draw a triangle on the Cartesian plane to evaluate $\dfrac{x}{r}$
- use a calculator
The reason for the answer is the subject of Question 2, below.

The value of $\cos 90°$ is $0$ .

Submit your answer as:
Question 1 was about the value of $\cos 90°$ . Which of the reasons below explains the answer to Question 1?
Answer:
The explanation for the answer to Question 1 is that when $\theta = 90°$ , .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use a picture of a triangle on the Cartesian plane with $\theta = 90°$ .

STEP: Use a triangle on the Cartesian plane to answer the question
[−1 point ⇒ 0 / 1 points left]

We need to identify the reason why $\cos 90°$ is $0$ . The answer comes directly from the definition for the cosine ratio on the Cartesian plane. That definition is:

$\begin{array}{c} \cos \theta = \dfrac{x}{r} \end{array}$

Remember that the ratio on the right refers to two sides of a triangle on the Cartesian plane:

The angle $\theta$ starts at the positive $x$ -axis and curves up to meet the hypotenuse of the triangle. Imagine that $\theta$ becomes $90°$ . When that happens, the hypotenuse will rotate up. And the triangle gets more thin until the hypotenuse and the vertical side swing up and over to meet on the $y$ -axis.

For $\theta = 90°$ , the triangle collapses to a vertical line segment on the $y$ -axis. The segment reaches from the origin up to the circle at the point $(0; y)$ . And the following things are both true about the "triangle" represented by that line:
- $x = 0$
- $y = r$
Let's use this information in the definition of the cosine ratio on the Cartesian plane:

$\begin{aligned} \cos \theta & = \dfrac{x}{r} \\ & \downarrow \\ \cos 90° & = \dfrac{0}{r} \\ & = 0 \end{aligned}$

And finally we can see why $\cos 90°$ must be $0$ : it is because when the angle is $90°$ , the side of the triangle corresponding to $x$ shrinks to zero.

The reason $\cos 90°$ is $0$ is because $x = 0$ .

Submit your answer as:

ID is: 3593 Seed is: 5507

Special angles: $0°$ and $90°$

On the Cartesian plane, the tangent ratio is defined as $\dfrac{y}{x}$ . Based on this definition or otherwise, determine the value of $\tan 90°$ .
INSTRUCTION:
- You may use a calculator.
- If the answer is undefined, type undefined.
Answer: The value of $\tan 90°$ is .
string
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you are not sure of the answer, use a calculator. That's right, we said it: use a calculator!

STEP: Evaluate the answer
[−1 point ⇒ 0 / 1 points left]

We need to determine the value of $\tan 90°$ . Here are two ways you might find the answer:
- draw a triangle on the Cartesian plane to evaluate $\dfrac{y}{x}$
- use a calculator
The reason for the answer is the subject of Question 2, below.

The value of $\tan 90°$ is undefined.

Submit your answer as:
Question 1 was about the value of $\tan 90°$ . Which of the reasons below explains the answer to Question 1?
Answer:
The explanation for the answer to Question 1 is that when $\theta = 90°$ , .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use a picture of a triangle on the Cartesian plane with $\theta = 90°$ .

STEP: Use a triangle on the Cartesian plane to answer the question
[−1 point ⇒ 0 / 1 points left]

We need to identify the reason why $\tan 90°$ is undefined. The answer comes directly from the definition for the tangent ratio on the Cartesian plane. That definition is:

$\begin{array}{c} \tan \theta = \dfrac{y}{x} \end{array}$

Remember that the ratio on the right refers to two sides of a triangle on the Cartesian plane:

The angle $\theta$ starts at the positive $x$ -axis and curves up to meet the hypotenuse of the triangle. Imagine that $\theta$ becomes $90°$ . When that happens, the hypotenuse will rotate up. And the triangle gets more thin until the hypotenuse and the vertical side swing up and over to meet on the $y$ -axis.

For $\theta = 90°$ , the triangle collapses to a vertical line segment on the $y$ -axis. The segment reaches from the origin up to the circle at the point $(0; y)$ . And the following things are both true about the "triangle" represented by that line:
- $x = 0$
- $y = r$
Let's use this information in the definition of the tangent ratio on the Cartesian plane:

$\begin{aligned} \tan \theta & = \dfrac{y}{x} \\ & \downarrow \\ \tan 90° & = \dfrac{y}{0} \\ & = \text{undefined} \end{aligned}$

And finally we can see why $\tan 90°$ must be undefined: it is because when the angle is $90°$ , the side of the triangle corresponding to $x$ shrinks to zero, leading to division by zero.

The reason $\tan 90°$ is undefined is because $x = 0$ .

Submit your answer as:

ID is: 3593 Seed is: 6870

Special angles: $0°$ and $90°$

On the Cartesian plane, the tangent ratio is defined as $\dfrac{y}{x}$ . Based on this definition or otherwise, determine the value of $\tan 90°$ .
INSTRUCTION:
- You may use a calculator.
- If the answer is undefined, type undefined.
Answer: The value of $\tan 90°$ is .
string
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you are not sure of the answer, use a calculator. That's right, we said it: use a calculator!

STEP: Evaluate the answer
[−1 point ⇒ 0 / 1 points left]

We need to determine the value of $\tan 90°$ . Here are two ways you might find the answer:
- draw a triangle on the Cartesian plane to evaluate $\dfrac{y}{x}$
- use a calculator
The reason for the answer is the subject of Question 2, below.

The value of $\tan 90°$ is undefined.

Submit your answer as:
Question 1 was about the value of $\tan 90°$ . Which of the reasons below explains the answer to Question 1?
Answer:
The explanation for the answer to Question 1 is that when $\theta = 90°$ , .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use a picture of a triangle on the Cartesian plane with $\theta = 90°$ .

STEP: Use a triangle on the Cartesian plane to answer the question
[−1 point ⇒ 0 / 1 points left]

We need to identify the reason why $\tan 90°$ is undefined. The answer comes directly from the definition for the tangent ratio on the Cartesian plane. That definition is:

$\begin{array}{c} \tan \theta = \dfrac{y}{x} \end{array}$

Remember that the ratio on the right refers to two sides of a triangle on the Cartesian plane:

The angle $\theta$ starts at the positive $x$ -axis and curves up to meet the hypotenuse of the triangle. Imagine that $\theta$ becomes $90°$ . When that happens, the hypotenuse will rotate up. And the triangle gets more thin until the hypotenuse and the vertical side swing up and over to meet on the $y$ -axis.

For $\theta = 90°$ , the triangle collapses to a vertical line segment on the $y$ -axis. The segment reaches from the origin up to the circle at the point $(0; y)$ . And the following things are both true about the "triangle" represented by that line:
- $x = 0$
- $y = r$
Let's use this information in the definition of the tangent ratio on the Cartesian plane:

$\begin{aligned} \tan \theta & = \dfrac{y}{x} \\ & \downarrow \\ \tan 90° & = \dfrac{y}{0} \\ & = \text{undefined} \end{aligned}$

And finally we can see why $\tan 90°$ must be undefined: it is because when the angle is $90°$ , the side of the triangle corresponding to $x$ shrinks to zero, leading to division by zero.

The reason $\tan 90°$ is undefined is because $x = 0$ .

Submit your answer as:

ID is: 3594 Seed is: 8043

Special angles: $180°$ , $270°$ , and $360°$

On the Cartesian plane, the tangent function is defined as $\dfrac{y}{x}$ . Based on this definition or otherwise, determine the value of $\tan 270°$ .
INSTRUCTION:
- You may use a calculator.
- If the answer is undefined, type undefined.
Answer: The value of $\tan 270°$ is .
string
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you are not sure of the answer, use a calculator. That's right, we said it: use a calculator!

STEP: Evaluate the answer
[−1 point ⇒ 0 / 1 points left]

We need to determine the value of $\tan 270°$ . Here are two ways you might find the answer:
- draw a triangle on the Cartesian plane to evaluate $\dfrac{y}{x}$
- use a calculator
The reason for the answer is the subject of Question 2, below.

The value of $\tan 270°$ is undefined.

Submit your answer as:
Question 1 was about the value of $\tan 270°$ . Which of the reasons below explains the answer to Question 1?
Answer: The explanation for the answer to Question 1 is that when $\theta = 270°$ , .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use a picture of a triangle on the Cartesian plane with $\theta = 270°$ .

STEP: Use a triangle on the Cartesian plane to answer the question
[−1 point ⇒ 0 / 1 points left]

We need to identify the reason why $\tan 270°$ is undefined. The answer comes directly from the definition for the tangent ratio on the Cartesian plane. That definition is:

$\begin{array}{c} \tan \theta = \dfrac{y}{x} \end{array}$

Remember that the ratio on the right refers to two sides of a triangle on the Cartesian plane:

The angle $\theta$ starts at the positive $x$ -axis and curves up to meet the hypotenuse of the triangle. Imagine that $\theta$ becomes $270°$ . When that happens, the hypotenuse will rotate into Quadrant II and then into Quadrant III, coming to a rest on the negative $y$ -axis (pointing straight down). This is shown below.

For $\theta = 270°$ , the triangle collapses to a vertical line segment on the $y$ -axis. The segment reaches from the origin down to the circle at the point $(0; y)$ . And the following things are both true about the "triangle" represented by that line:
- $x = 0$
- $y = -r$
Remember that $r$ , the radius, is positive. But for $\theta = 270°$ we know that $y$ must be negative (you can see this on the diagram above). That is the reason why $y = -r$ , not $y = r$ .

Now we can use this information in the definition of the tangent ratio on the Cartesian plane:

$\begin{aligned} \tan \theta & = \dfrac{y}{x} \\ & \downarrow \\ \tan 270° & = \dfrac{y}{0} \\ & = \text{undefined} \end{aligned}$

And finally we can see why $\tan 270°$ must be undefined: it is because when the angle is $270°$ , the side of the triangle corresponding to $x$ shrinks to zero, leading to division by zero.

The reason $\tan 270°$ is undefined is because $x = 0$ .

Submit your answer as:

ID is: 3594 Seed is: 2030

Special angles: $180°$ , $270°$ , and $360°$

On the Cartesian plane, the cosine function is defined as $\dfrac{x}{r}$ . Based on this definition or otherwise, determine the value of $\cos 270°$ .
INSTRUCTION:
- You may use a calculator.
- If the answer is undefined, type undefined.
Answer: The value of $\cos 270°$ is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you are not sure of the answer, use a calculator. That's right, we said it: use a calculator!

STEP: Evaluate the answer
[−1 point ⇒ 0 / 1 points left]

We need to determine the value of $\cos 270°$ . Here are two ways you might find the answer:
- draw a triangle on the Cartesian plane to evaluate $\dfrac{x}{r}$
- use a calculator
The reason for the answer is the subject of Question 2, below.

The value of $\cos 270°$ is $0$ .

Submit your answer as:
Question 1 was about the value of $\cos 270°$ . Which of the reasons below explains the answer to Question 1?
Answer: The explanation for the answer to Question 1 is that when $\theta = 270°$ , .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use a picture of a triangle on the Cartesian plane with $\theta = 270°$ .

STEP: Use a triangle on the Cartesian plane to answer the question
[−1 point ⇒ 0 / 1 points left]

We need to identify the reason why $\cos 270°$ is $0$ . The answer comes directly from the definition for the cosine ratio on the Cartesian plane. That definition is:

$\begin{array}{c} \cos \theta = \dfrac{x}{r} \end{array}$

Remember that the ratio on the right refers to two sides of a triangle on the Cartesian plane:

The angle $\theta$ starts at the positive $x$ -axis and curves up to meet the hypotenuse of the triangle. Imagine that $\theta$ becomes $270°$ . When that happens, the hypotenuse will rotate into Quadrant II and then into Quadrant III, coming to a rest on the negative $y$ -axis (pointing straight down). This is shown below.

For $\theta = 270°$ , the triangle collapses to a vertical line segment on the $y$ -axis. The segment reaches from the origin down to the circle at the point $(0; y)$ . And the following things are both true about the "triangle" represented by that line:
- $x = 0$
- $y = -r$
Remember that $r$ , the radius, is positive. But for $\theta = 270°$ we know that $y$ must be negative (you can see this on the diagram above). That is the reason why $y = -r$ , not $y = r$ .

Now we can use this information in the definition of the cosine ratio on the Cartesian plane:

$\begin{aligned} \cos \theta & = \dfrac{x}{r} \\ & \downarrow \\ \cos 270° & = \dfrac{0}{r} \\ & = 0 \end{aligned}$

And finally we can see why $\cos 270°$ must be $0$ : it is because when the angle is $270°$ , the side of the triangle corresponding to $x$ shrinks to zero.

The reason $\cos 270°$ is $0$ is because $x = 0$ .

Submit your answer as:

ID is: 3594 Seed is: 8587

Special angles: $180°$ , $270°$ , and $360°$

On the Cartesian plane, the tangent function is defined as $\dfrac{y}{x}$ . Based on this definition or otherwise, determine the value of $\tan 270°$ .
INSTRUCTION:
- You may use a calculator.
- If the answer is undefined, type undefined.
Answer: The value of $\tan 270°$ is .
string
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

If you are not sure of the answer, use a calculator. That's right, we said it: use a calculator!

STEP: Evaluate the answer
[−1 point ⇒ 0 / 1 points left]

We need to determine the value of $\tan 270°$ . Here are two ways you might find the answer:
- draw a triangle on the Cartesian plane to evaluate $\dfrac{y}{x}$
- use a calculator
The reason for the answer is the subject of Question 2, below.

The value of $\tan 270°$ is undefined.

Submit your answer as:
Question 1 was about the value of $\tan 270°$ . Which of the reasons below explains the answer to Question 1?
Answer: The explanation for the answer to Question 1 is that when $\theta = 270°$ , .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use a picture of a triangle on the Cartesian plane with $\theta = 270°$ .

STEP: Use a triangle on the Cartesian plane to answer the question
[−1 point ⇒ 0 / 1 points left]

We need to identify the reason why $\tan 270°$ is undefined. The answer comes directly from the definition for the tangent ratio on the Cartesian plane. That definition is:

$\begin{array}{c} \tan \theta = \dfrac{y}{x} \end{array}$

Remember that the ratio on the right refers to two sides of a triangle on the Cartesian plane:

The angle $\theta$ starts at the positive $x$ -axis and curves up to meet the hypotenuse of the triangle. Imagine that $\theta$ becomes $270°$ . When that happens, the hypotenuse will rotate into Quadrant II and then into Quadrant III, coming to a rest on the negative $y$ -axis (pointing straight down). This is shown below.

For $\theta = 270°$ , the triangle collapses to a vertical line segment on the $y$ -axis. The segment reaches from the origin down to the circle at the point $(0; y)$ . And the following things are both true about the "triangle" represented by that line:
- $x = 0$
- $y = -r$
Remember that $r$ , the radius, is positive. But for $\theta = 270°$ we know that $y$ must be negative (you can see this on the diagram above). That is the reason why $y = -r$ , not $y = r$ .

Now we can use this information in the definition of the tangent ratio on the Cartesian plane:

$\begin{aligned} \tan \theta & = \dfrac{y}{x} \\ & \downarrow \\ \tan 270° & = \dfrac{y}{0} \\ & = \text{undefined} \end{aligned}$

And finally we can see why $\tan 270°$ must be undefined: it is because when the angle is $270°$ , the side of the triangle corresponding to $x$ shrinks to zero, leading to division by zero.

The reason $\tan 270°$ is undefined is because $x = 0$ .

Submit your answer as:

ID is: 3654 Seed is: 3333

Trigonometry gone wild!

Point $K$ is given on the Cartesian plane with the origin at $O$ . It is in the first quadrant at $\left( 14 ; y_{K} \right)$ , where $y_{K}$ is the $y$ -coordinate of Point $K$ . $K$ makes an angle $\phi$ with the positive $x$ -axis, as labelled, and $\overline{OK}$ is $\text{50}$ units long. Given that $\tan \phi = \dfrac{24}{7}$ , determine the value of $y_{K}$ . The diagram may or may not be drawn to scale.

Answer: $y_{K}$ =
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the definition of the tangent ratio on the Cartesian plane to write an equation which includes the $y$ -coordinate.

STEP: Write an equation based on the information given
[−1 point ⇒ 1 / 2 points left]

We need to determine the $y$ -coordinate of Point $K$ . The point sits at an angle of $\phi$ to the positive $x$ -axis and we know that $\tan \phi$ is equal to $\dfrac{24}{7}$ . As always, the tangent ratio is defined as $\dfrac{y}{x}$ . So we can write:

$\begin{array}{c}\dfrac{y}{x} = \dfrac{24}{7}\end{array}$

This equation is true for the angle $\phi$ and Point $K$ . So we can substitute in the $x$ -coordinate value of $\text{14}$ on the left-hand side of the equation. (We will also change $y$ to $y_{K}$ because the equation is specifically about Point $K$ now.)

$\begin{array}{c}\dfrac{y_{K}}{\left(14\right)} = \dfrac{24}{7}\end{array}$

NOTE:
The equation above is a proportion. It represents the fact that any point at an angle of $\phi$ will have the same ratio value, no matter how far the point is from the origin. We can see this best on the Cartesian plane. Here is the same figure as in the question, but with a second point which is also at an angle $\phi$ . The second point makes a smaller triangle.

The triangles shown above are similar. And that is where the proportion comes from: similar shapes have proportional sides. The trigonometric ratios are built on similarity. Trigonometry does not care how big a triangle is - it only cares how big the angles are because that determines the ratio of the triangle's sides!

STEP: Solve the equation
[−1 point ⇒ 0 / 2 points left]

Now we need to solve the equation. Multiply both sides by the denominator on the left side to isolate $y_K$ .

$\begin{aligned} \dfrac{y_{K}}{\left(14\right)} & = \dfrac{24}{7} \\ y_{K} & = \dfrac{24}{7} \left( 14 \right)\\ y_{K} & = 48 \end{aligned}$

TIP: Check the sign of the answer: it must be positive because Point $K$ is in the first quadrant. If your sign is wrong, you should go back and check through your work.

The $y$ -coordinate of Point $K$ is $\text{48}$ .

Submit your answer as:
Evaluate the following expression:

$\begin{aligned} - \dfrac{25}{96}\sin ^2\phi+1 \end{aligned}$

INSTRUCTION: Your answer must be a simplified fraction.

Answer: $- \dfrac{25}{96}\sin ^2\phi+1 =$
fraction
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the result of Question 1 to calculate the value of $\sin \phi$ . Then evaluate the expression.

STEP: Determine the value of $\sin \phi$
[−1 point ⇒ 1 / 2 points left]

We can start by finding the value of $\sin \phi$ . From Question 1 we know everything about Point $K$ (which is linked to the angle $\phi$ ):

On the Cartesian plane the sine ratio is $\dfrac{y}{r}$ . Using the values from the figure above we get:

$\begin{aligned} \sin \theta & = \dfrac{y}{r} \\ \sin \phi & = \frac{\left(48\right)}{\left(50\right)} \\ & = \dfrac{24}{25} \end{aligned}$

STEP: Evaluate the expression
[−1 point ⇒ 0 / 2 points left]

Now we can evaluate the expression. Substitute $\dfrac{24}{25}$ in for $\sin \phi$ and simplify:

$\begin{aligned} & - \dfrac{25}{96}\sin ^2\phi+1 \\ & = - \frac{25}{96} \left( \frac{24}{25} \right) ^ { 2 } + 1\\ &=- \frac{25}{96} \left( \frac{576}{625} \right) + 1\\ &=- \frac{6}{25} + 1\\ &=-\frac{6}{25} + \frac{25}{25}\\ &=\frac{19}{25} \end{aligned}$

The value of the expression is $\dfrac{19}{25}$ .

Submit your answer as:

ID is: 3654 Seed is: 601

Trigonometry gone wild!

Point $K$ is given on the Cartesian plane with the origin at $O$ . It is in the first quadrant at $\left( 30 ; y_{K} \right)$ , where $y_{K}$ is the $y$ -coordinate of Point $K$ . $K$ makes an angle $\phi$ with the positive $x$ -axis, as labelled, and $\overline{OK}$ is $\text{34}$ units long. Given that $\sin \phi = \dfrac{8}{17}$ , determine the value of $y_{K}$ . The diagram may or may not be drawn to scale.

Answer: $y_{K}$ =
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the definition of the sine ratio on the Cartesian plane to write an equation which includes the $y$ -coordinate.

STEP: Write an equation based on the information given
[−1 point ⇒ 1 / 2 points left]

We need to determine the $y$ -coordinate of Point $K$ . The point sits at an angle of $\phi$ to the positive $x$ -axis and we know that $\sin \phi$ is equal to $\dfrac{8}{17}$ . As always, the sine ratio is defined as $\dfrac{y}{r}$ . So we can write:

$\begin{array}{c}\dfrac{y}{r} = \dfrac{8}{17}\end{array}$

This equation is true for the angle $\phi$ and Point $K$ . So we can substitute in the radius value of $\text{34}$ on the left-hand side of the equation. (We will also change $y$ to $y_{K}$ because the equation is specifically about Point $K$ now.)

$\begin{array}{c}\dfrac{y_{K}}{\left(34\right)} = \dfrac{8}{17}\end{array}$

NOTE:
The equation above is a proportion. It represents the fact that any point at an angle of $\phi$ will have the same ratio value, no matter how far the point is from the origin. We can see this best on the Cartesian plane. Here is the same figure as in the question, but with a second point which is also at an angle $\phi$ . The second point makes a smaller triangle.

The triangles shown above are similar. And that is where the proportion comes from: similar shapes have proportional sides. The trigonometric ratios are built on similarity. Trigonometry does not care how big a triangle is - it only cares how big the angles are because that determines the ratio of the triangle's sides!

STEP: Solve the equation
[−1 point ⇒ 0 / 2 points left]

Now we need to solve the equation. Multiply both sides by the denominator on the left side to isolate $y_K$ .

$\begin{aligned} \dfrac{y_{K}}{\left(34\right)} & = \dfrac{8}{17} \\ y_{K} & = \dfrac{8}{17} \left( 34 \right)\\ y_{K} & = 16 \end{aligned}$

TIP: Check the sign of the answer: it must be positive because Point $K$ is in the first quadrant. If your sign is wrong, you should go back and check through your work.

The $y$ -coordinate of Point $K$ is $\text{16}$ .

Submit your answer as:
Evaluate the following expression:

$\begin{aligned} \dfrac{15}{4}\tan ^2\phi-5 \end{aligned}$

INSTRUCTION: Your answer must be a simplified fraction.

Answer: $\dfrac{15}{4}\tan ^2\phi-5 =$
fraction
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the result of Question 1 to calculate the value of $\tan \phi$ . Then evaluate the expression.

STEP: Determine the value of $\tan \phi$
[−1 point ⇒ 1 / 2 points left]

We can start by finding the value of $\tan \phi$ . From Question 1 we know everything about Point $K$ (which is linked to the angle $\phi$ ):

On the Cartesian plane the tangent ratio is $\dfrac{y}{x}$ . Using the values from the figure above we get:

$\begin{aligned} \tan \theta & = \dfrac{y}{x} \\ \tan \phi & = \frac{\left(16\right)}{\left(30\right)} \\ & = \dfrac{8}{15} \end{aligned}$

STEP: Evaluate the expression
[−1 point ⇒ 0 / 2 points left]

Now we can evaluate the expression. Substitute $\dfrac{8}{15}$ in for $\tan \phi$ and simplify:

$\begin{aligned} & \dfrac{15}{4}\tan ^2\phi-5 \\ & = \frac{15}{4} \left( \frac{8}{15} \right) ^ { 2 } - 5\\ &=\frac{15}{4} \left( \frac{64}{225} \right) - 5\\ &=\frac{16}{15} - 5\\ &=\frac{16}{15} - \frac{75}{15}\\ &=- \frac{59}{15} \end{aligned}$

The value of the expression is $- \dfrac{59}{15}$ .

Submit your answer as:

ID is: 3654 Seed is: 6185

Trigonometry gone wild!

Point $L$ is given on the Cartesian plane with the origin at $O$ . It is in the second quadrant at $\left( x_{L} ; 18 \right)$ , where $x_{L}$ is the $x$ -coordinate of Point $L$ . $L$ makes an angle $\phi$ with the positive $x$ -axis, as labelled, and $\overline{OL}$ is $\text{30}$ units long. Given that $\tan \phi = - \dfrac{3}{4}$ , determine the value of $x_{L}$ . The diagram may or may not be drawn to scale.

Answer: $x_{L}$ =
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the definition of the tangent ratio on the Cartesian plane to write an equation which includes the $x$ -coordinate.

STEP: Write an equation based on the information given
[−1 point ⇒ 1 / 2 points left]

We need to determine the $x$ -coordinate of Point $L$ . The point sits at an angle of $\phi$ to the positive $x$ -axis and we know that $\tan \phi$ is equal to $- \dfrac{3}{4}$ . As always, the tangent ratio is defined as $\dfrac{y}{x}$ . So we can write:

$\begin{array}{c}\dfrac{y}{x} = - \dfrac{3}{4}\end{array}$

This equation is true for the angle $\phi$ and Point $L$ . So we can substitute in the $y$ -coordinate value of $\text{18}$ on the left-hand side of the equation. (We will also change $x$ to $x_{L}$ because the equation is specifically about Point $L$ now.)

$\begin{array}{c}\dfrac{\left(18\right)}{x_{L}} = - \dfrac{3}{4}\end{array}$

NOTE:
The equation above is a proportion. It represents the fact that any point at an angle of $\phi$ will have the same ratio value, no matter how far the point is from the origin. We can see this best on the Cartesian plane. Here is the same figure as in the question, but with a second point which is also at an angle $\phi$ . The second point makes a smaller triangle.

The triangles shown above are similar. And that is where the proportion comes from: similar shapes have proportional sides. The trigonometric ratios are built on similarity. Trigonometry does not care how big a triangle is - it only cares how big the angles are because that determines the ratio of the triangle's sides!

STEP: Solve the equation
[−1 point ⇒ 0 / 2 points left]

Now we need to solve the equation. Start by taking the reciprocal of both sides to get $x_L$ out of the demoninator. Then multiply both sides by the denominator on the left side to isolate $x_L$ .

$\begin{aligned} \dfrac{\left(18\right)}{x_{L}} & = - \dfrac{3}{4} \\ \frac{x_{L}}{ 18 } & = - \frac{4}{3}\\ x_{L} & = - \frac{4}{3} \left( 18 \right)\\ x_{L} & = -24 \end{aligned}$

TIP: Check the sign of the answer: it must be negative because Point $L$ is in the second quadrant. If your sign is wrong, you should go back and check through your work.

The $x$ -coordinate of Point $L$ is $\text{−24}$ .

Submit your answer as:
Evaluate the following expression:

$\begin{aligned} \dfrac{25}{6}\cos \phi+2 \end{aligned}$

INSTRUCTION: Your answer must be a simplified fraction.

Answer: $\dfrac{25}{6}\cos \phi+2 =$
fraction
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the result of Question 1 to calculate the value of $\cos \phi$ . Then evaluate the expression.

STEP: Determine the value of $\cos \phi$
[−1 point ⇒ 1 / 2 points left]

We can start by finding the value of $\cos \phi$ . From Question 1 we know everything about Point $L$ (which is linked to the angle $\phi$ ):

On the Cartesian plane the cosine ratio is $\dfrac{x}{r}$ . Using the values from the figure above we get:

$\begin{aligned} \cos \theta & = \dfrac{x}{r} \\ \cos \phi & = \frac{\left(-24\right)}{\left(30\right)} \\ & = - \dfrac{4}{5} \end{aligned}$

STEP: Evaluate the expression
[−1 point ⇒ 0 / 2 points left]

Now we can evaluate the expression. Substitute $- \dfrac{4}{5}$ in for $\cos \phi$ and simplify:

$\begin{aligned} & \dfrac{25}{6}\cos \phi+2 \\ & = \frac{25}{6} \left( - \frac{4}{5} \right) + 2\\ &=- \frac{10}{3} + 2\\ &=-\frac{10}{3} + \frac{6}{3}\\ &=- \frac{4}{3} \end{aligned}$

The value of the expression is $- \dfrac{4}{3}$ .

Submit your answer as:

ID is: 3603 Seed is: 5196

Getting familiar with the CAST diagram

The Cartesian plane below can be used for the CAST diagram. The quadrants (I, II, III, and IV) are labelled.

In the CAST diagram, the letters C, A, S, and T each belong in one quadrant. Where on the Cartesian plane does the letter C belong?

Answer:

The C should be in Quadrant .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find a section on the CAST diagram in the Everything Maths textbook.

STEP: Add the letters to the Cartesian plane
[−1 point ⇒ 0 / 1 points left]

This question is about something very useful in trigonometry: the CAST diagram. Each letter in CAST refers to trigonometric ratios:

C - cosine
A - all (sine, cosine, and tangent)
S - sine
T - tangent

The letters sit on the Cartesian plane as shown below: C is in the lower right quadrant (Quadrant IV). The letters C-A-S-T go around the diagram in the anticlockwise direction. You will find it helpful to memorise the positions of the letters!

The C belongs in Quadrant IV. The C in Quadrant IV tells us that the cosine ratio is positive for any angle in Quadrant IV. At the same time, it tells us that the sine and tangent ratios are negative in that quadrant.

Here are some useful facts about the CAST diagram:

In Quadrant I all the trigonometric ratios are positive. This means all three ratios are positive for angles between $0°$ and $90°$ .
Each trigonometric ratio is positive in $\text{2}$ of the quadrants and negative in $\text{2}$ of the quadrants. For example, the cosine ratio is positive in Quadrants I and IV, while it is negative in Quadrants II and III.

The C should be in Quadrant IV.

Submit your answer as:

ID is: 3603 Seed is: 2506

Getting familiar with the CAST diagram

The Cartesian plane below can be used for the CAST diagram. The quadrants (I, II, III, and IV) are labelled.

In the CAST diagram, the letters C, A, S, and T each belong in one quadrant. Where on the Cartesian plane does the letter T belong?

Answer:

The T should be in Quadrant .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find a section on the CAST diagram in the Everything Maths textbook.

STEP: Add the letters to the Cartesian plane
[−1 point ⇒ 0 / 1 points left]

This question is about something very useful in trigonometry: the CAST diagram. Each letter in CAST refers to trigonometric ratios:

C - cosine
A - all (sine, cosine, and tangent)
S - sine
T - tangent

The T belongs in Quadrant III. The T in Quadrant III tells us that the tangent ratio is positive for any angle in Quadrant III. At the same time, it tells us that the sine and cosine ratios are negative in that quadrant.

Here are some useful facts about the CAST diagram:

In Quadrant I all the trigonometric ratios are positive. This means all three ratios are positive for angles between $0°$ and $90°$ .
Each trigonometric ratio is positive in $\text{2}$ of the quadrants and negative in $\text{2}$ of the quadrants. For example, the cosine ratio is positive in Quadrants I and IV, while it is negative in Quadrants II and III.

The T should be in Quadrant III.

Submit your answer as:

ID is: 3603 Seed is: 5898

Getting familiar with the CAST diagram

The Cartesian plane below can be used for the CAST diagram. The quadrants (I, II, III, and IV) are labelled.

In the CAST diagram, the letters C, A, S, and T each belong in one quadrant. Where on the Cartesian plane does the letter A belong?

Answer:

The A should be in Quadrant .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find a section on the CAST diagram in the Everything Maths textbook.

STEP: Add the letters to the Cartesian plane
[−1 point ⇒ 0 / 1 points left]

This question is about something very useful in trigonometry: the CAST diagram. Each letter in CAST refers to trigonometric ratios:

C - cosine
A - all (sine, cosine, and tangent)
S - sine
T - tangent

The A belongs in Quadrant I. The A in Quadrant I tells us that all three of the trigonometric ratios are positive for any angle in Quadrant I (any angle between $0°$ and $90°$ ).

Here are some useful facts about the CAST diagram:

In Quadrant I all the trigonometric ratios are positive. This means all three ratios are positive for angles between $0°$ and $90°$ .
Each trigonometric ratio is positive in $\text{2}$ of the quadrants and negative in $\text{2}$ of the quadrants. For example, the cosine ratio is positive in Quadrants I and IV, while it is negative in Quadrants II and III.

The A should be in Quadrant I.

Submit your answer as:

ID is: 3655 Seed is: 6410

Transferring trigonometric ratios to the Cartesian plane

Consider the following equations about an angle $\theta$ :

\begin{aligned} \sin θ & = \frac{5}{13} \\ \cos θ & = \frac{12}{13} \end{aligned}

$\begin{aligned} \sin \theta & = \frac{5}{13} \\[0.4cm] \cos \theta & = \frac{12}{13} \end{aligned}$

We can draw a reference triangle on the Cartesian plane to represent the angle $\theta$ . One vertex of the triangle will be in Quadrant I, II, III, or IV. What are the coordinates of this point?

INSTRUCTION: Type your answer as a coordinate pair with brackets, like this: (2 ; -6).

Answer: The coordinates are: .

coordinate

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The first equation shows that the sine ratio is positive and the second equation shows that the cosine ratio is positive. Use these two facts and the CAST diagram to figure out which quadrant the angle $\theta$ must point into.

STEP: Determine which quadrant the angle points into
[−1 point ⇒ 3 / 4 points left]

We need to determine the coordinates of the point on the Cartesian plane which correspond to the reference triangle for the angle $\theta$ . The first thing we need to do is figure out which quadrant the angle points into. Then we can draw a point and the reference triangle we need in that quadrant.

We can use the signs of the two ratios given and the CAST diagram to figure out which quadrant the angle $\theta$ must point into. The first equation shows that the sine ratio is positive and the second equation shows that the cosine ratio is positive. So the point must be in Quadrant I: that is the only quadrant where the sine and the cosine ratios are both positive.

STEP: Draw a sketch using the CAST diagram
[−1 point ⇒ 2 / 4 points left]

Now we can draw a point in Quadrant I, with a reference triangle. We do not know exactly where the point is, so we can just pick a point somewhere in Quadrant I.

NOTE: This diagram is a sketch, so it is not to scale.

STEP: Read the values of

x

$x$ and

y

$y$ from the given ratios
[−2 points ⇒ 0 / 4 points left]

Now we can use the two equations given to find the coordinates of the point. We do this using the definitions of the sine and cosine ratios.

The definition for sine is $\dfrac{y}{r}$ . The $y$ in this definition is one of the coordinates we want!

\begin{aligned} \sin θ = \frac{5}{13} ⟺ & \frac{y}{r} \\ which means: \\ y & = 5 \\ r & = 13 \end{aligned}

$\begin{aligned} \sin \theta = \frac{5}{13} \quad \Longleftrightarrow \quad & \dfrac{y}{r} \\ \text{which means:} \quad & \\ y & = 5 \\ r & = 13 \end{aligned}$

The definition for cosine is $\dfrac{x}{r}$ . And we can follow the same logic as above to find the value of $x$ .

\begin{aligned} \cos θ = \frac{12}{13} ⟺ & \frac{x}{r} \\ which means: \\ x & = 12 \\ r & = 13 \end{aligned}

$\begin{aligned} \cos \theta = \frac{12}{13} \quad \Longleftrightarrow \quad & \dfrac{x}{r} \\ \text{which means:} \quad & \\ x & = 12 \\ r & = 13 \end{aligned}$

Now we know that the coordinates of the point in the reference triangle are $\left(12; 5\right)$ . And we also know the radius, $r$ , for the reference triangle. The reference triangle looks like this:

NOTE: While the lengths of the triangle's sides must be positive, the coordinates of the point can be positive or negative, depending on which quadrant the point is in. It is crucial to be aware that these signs can be different.

The coordinates of the point in the reference triangle are $\left(12; 5\right)$ .

Submit your answer as:

ID is: 3655 Seed is: 8140

Transferring trigonometric ratios to the Cartesian plane

Consider the following equations about an angle $\theta$ :

\begin{aligned} \tan θ & = \frac{12}{5} \\ \cos θ & = \frac{5}{13} \end{aligned}

$\begin{aligned} \tan \theta & = \frac{12}{5} \\[0.4cm] \cos \theta & = \frac{5}{13} \end{aligned}$

We can draw a reference triangle on the Cartesian plane to represent the angle $\theta$ . One vertex of the triangle will be in Quadrant I, II, III, or IV. What are the coordinates of this point?

INSTRUCTION: Type your answer as a coordinate pair with brackets, like this: (2 ; -6).

Answer: The coordinates are: .

coordinate

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The first equation shows that the tangent ratio is positive and the second equation shows that the cosine ratio is positive. Use these two facts and the CAST diagram to figure out which quadrant the angle $\theta$ must point into.

STEP: Determine which quadrant the angle points into
[−1 point ⇒ 3 / 4 points left]

We can use the signs of the two ratios given and the CAST diagram to figure out which quadrant the angle $\theta$ must point into. The first equation shows that the tangent ratio is positive and the second equation shows that the cosine ratio is positive. So the point must be in Quadrant I: that is the only quadrant where the tangent and the cosine ratios are both positive.

STEP: Draw a sketch using the CAST diagram
[−1 point ⇒ 2 / 4 points left]

Now we can draw a point in Quadrant I, with a reference triangle. We do not know exactly where the point is, so we can just pick a point somewhere in Quadrant I.

NOTE: This diagram is a sketch, so it is not to scale.

STEP: Read the values of

x

$x$ and

y

$y$ from the given ratios
[−2 points ⇒ 0 / 4 points left]

Now we can use the two equations given to find the coordinates of the point. We do this using the definitions of the tangent and cosine ratios.

In fact, the tangent ratio is enough to tell us the coordinates. This is because the tangent ratio is always equal to $\dfrac{y}{x}$ . So we can pull the values of $x$ and $y$ out of this equation:

\begin{matrix} \tan θ = \frac{12}{5} ⟺ \frac{y}{x} \end{matrix}

$\begin{array}{c}\tan \theta = \frac{12}{5} \quad \Longleftrightarrow \quad \frac{y}{x}\end{array}$

The ratio is positive. But that does not mean that $x$ and $y$ are both positive: they might both be negative (because a negative divided by a negative makes a positive). So either $x$ and $y$ are both positive, or they are both negative. Which is it? The answer comes from the fact that we already know which quadrant the point is in! The point is in Quadrant I, which means both of the coordinates must be positive. Therefore:

\begin{aligned} x & = 5 \\ y & = 12 \end{aligned}

$\begin{aligned} x & = 5 \\ y & = 12 \end{aligned}$

Now we know that the coordinates of the point in the reference triangle are $\left(5; 12\right)$ . Note that you can also read the radius value (the hypotenuse of the triangle) from the cosine ratio, which is equal to $\frac{5}{13}$ . The radius is equal to $\text{13}$ . The reference triangle looks like this:

The coordinates of the point in the reference triangle are $\left(5; 12\right)$ .

Submit your answer as:

ID is: 3655 Seed is: 2846

Transferring trigonometric ratios to the Cartesian plane

Consider the following equations about an angle $\theta$ :

\begin{aligned} \tan θ & = - \frac{12}{5} \\ \cos θ & = \frac{5}{13} \end{aligned}

$\begin{aligned} \tan \theta & = - \frac{12}{5} \\[0.4cm] \cos \theta & = \frac{5}{13} \end{aligned}$

We can draw a reference triangle on the Cartesian plane to represent the angle $\theta$ . One vertex of the triangle will be in Quadrant I, II, III, or IV. What are the coordinates of this point?

INSTRUCTION: Type your answer as a coordinate pair with brackets, like this: (2 ; -6).

Answer: The coordinates are: .

coordinate

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The first equation shows that the tangent ratio is negative and the second equation shows that the cosine ratio is positive. Use these two facts and the CAST diagram to figure out which quadrant the angle $\theta$ must point into.

STEP: Determine which quadrant the angle points into
[−1 point ⇒ 3 / 4 points left]

We can use the signs of the two ratios given and the CAST diagram to figure out which quadrant the angle $\theta$ must point into. The first equation shows that the tangent ratio is negative and the second equation shows that the cosine ratio is positive. So the point must be in Quadrant IV: that is the only quadrant where the tangent ratio is negative and the cosine ratio is positive.

STEP: Draw a sketch using the CAST diagram
[−1 point ⇒ 2 / 4 points left]

Now we can draw a point in Quadrant IV, with a reference triangle. We do not know exactly where the point is, so we can just pick a point somewhere in Quadrant IV.

NOTE: This diagram is a sketch, so it is not to scale.

STEP: Read the values of

x

$x$ and

y

$y$ from the given ratios
[−2 points ⇒ 0 / 4 points left]

Now we can use the two equations given to find the coordinates of the point. We do this using the definitions of the tangent and cosine ratios.

In fact, the tangent ratio is enough to tell us the coordinates. This is because the tangent ratio is always equal to $\dfrac{y}{x}$ . So we can pull the values of $x$ and $y$ out of this equation:

\begin{matrix} \tan θ = - \frac{12}{5} ⟺ \frac{y}{x} \end{matrix}

$\begin{array}{c}\tan \theta = - \frac{12}{5} \quad \Longleftrightarrow \quad \frac{y}{x}\end{array}$

We can see that the ratio is negative. That means either $x$ or $y$ is negative (only one of them, not both!). So which one is negative? The answer comes from the fact that we already know which quadrant the point is in! The point is in Quadrant IV. So the $x$ -coordinate must be positive and the $y$ -coordinate must be negative. Therefore:

\begin{aligned} x & = 5 \\ y & = - 12 \end{aligned}

$\begin{aligned} x & = 5 \\ y & = -12 \end{aligned}$

Now we know that the coordinates of the point in the reference triangle are $\left(5; -12\right)$ . Note that you can also read the radius value (the hypotenuse of the triangle) from the cosine ratio, which is equal to $\frac{5}{13}$ . The radius is equal to $\text{13}$ . The reference triangle looks like this:

The coordinates of the point in the reference triangle are $\left(5; -12\right)$ .

Submit your answer as:

ID is: 3653 Seed is: 2834

Trigonometric ratios on the Cartesian plane: positives & negatives

Consider the following equation about an angle $\theta$ :

$\begin{array}{c}\tan \theta = - \dfrac{8}{15}\end{array}$

The diagram below shows four points on the Cartesian plane. They are labelled $W, X, Y$ , and $Z$ . The diagram is not drawn to scale.

Two of the four points above correspond to the equation $\tan \theta = - \dfrac{8}{15}$ . Which two?

Answer:
The points corresponding to $\tan \theta = - \dfrac{8}{15}$ are Points .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to compare the sign of the ratio to the CAST diagram.

STEP: Use the CAST diagram and the sign of the ratio given
[−1 point ⇒ 0 / 1 points left]

On the Cartesian plane, there is a connection between the sign of the trigonometric ratios and which quadrant the angle points into. For each of the ratios, there are two quadrants where the ratio is positive and two quadrants where the ratio is negative. The CAST diagram shows us these quadrants. So we can use the CAST diagram to answer this question.

In this case we know that:

$\begin{aligned} \tan \theta = & \underbrace{- \dfrac{8}{15}} \\ & \hspace{-0.4cm}\text{negative} \end{aligned}$

This equation tells us that the angle $\theta$ must point into one of the two quadrants where the tangent ratio is negative. Let's add the CAST diagram letters onto the diagram given in the question to find those two quadrants. Each letter of the CAST diagram tells us which trigonometric ratio is positive in that quadrant.

We are dealing with the tangent ratio, and we know that the ratio has a negative value. The A in Quadrant I tells us that all three ratios are positive in that quadrant. So we do not want Point $W$ . Similarly, the T in Quadrant III means that the tangent ratio is positive in that quadrant so we do not want Point $Y$ either. So Points $X$ and $Z$ are the points which correspond to the equation in the question.

The correct points are $X$ and $Z$ .

Submit your answer as:
In Question 1 you found the two points which correspond to this equation:

$\begin{array}{c}\tan \theta = - \dfrac{8}{15}\end{array}$

There are two values for $\theta$ connected to those points. From the list of choices below, select the two values closest to $\theta$ (the choices in the list have been rounded).

TIP: Use the answer to Question 1.

Answer: The values for $\theta$ are closest to: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to pick the angles which match the correct points in Question 1.

STEP: Use the answer to Question 1
[−1 point ⇒ 0 / 1 points left]

For this question we need to identify the values of $\theta$ which correspond to the equation

$\begin{array}{c}\tan \theta = - \dfrac{8}{15}\end{array}$

There are two such values: they match the points we identified in Question 1.

We already know that Points $X$ and $Z$ correspond to the equation. So we need to find the angles which match those points. The choices in the list include these four angles: $28° ; 152° ; 208° ; 332°$ . Two of these choices agree with the positions of Points $X$ and $Z$ . (Remember that the angles always start on the positive $x$ -axis, as shown below.)

We do not know the exact values of these two angles. But we know that one of them points into Quadrant II and the other points into Quadrant IV. That means the smaller angle must be obtuse (between $\text{90}\text{°}$ and $\text{180}\text{°}$ ) and the larger angle must be between $\text{270}\text{°}$ and $\text{360}\text{°}$ . Based on the choices in the list, the values for $\theta$ must be $152°$ and $332°$ .

The two values for $\theta$ are $152°$ and $332°$ .

Submit your answer as:

ID is: 3653 Seed is: 416

Trigonometric ratios on the Cartesian plane: positives & negatives

Consider the following equation about an angle $\theta$ :

$\begin{array}{c}\tan \theta = \dfrac{5}{12}\end{array}$

The diagram below shows four points on the Cartesian plane. They are labelled $W, X, Y$ , and $Z$ . The diagram is not drawn to scale.

Two of the four points above correspond to the equation $\tan \theta = \dfrac{5}{12}$ . Which two?

Answer:
The points corresponding to $\tan \theta = \dfrac{5}{12}$ are Points .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to compare the sign of the ratio to the CAST diagram.

STEP: Use the CAST diagram and the sign of the ratio given
[−1 point ⇒ 0 / 1 points left]

On the Cartesian plane, there is a connection between the sign of the trigonometric ratios and which quadrant the angle points into. For each of the ratios, there are two quadrants where the ratio is positive and two quadrants where the ratio is negative. The CAST diagram shows us these quadrants. So we can use the CAST diagram to answer this question.

In this case we know that:

$\begin{aligned} \tan \theta = & \underbrace{\dfrac{5}{12}} \\ & \hspace{-0.4cm}\text{positive} \end{aligned}$

This equation tells us that the angle $\theta$ must point into one of the two quadrants where the tangent ratio is positive. Let's add the CAST diagram letters onto the diagram given in the question to find those two quadrants. Each letter of the CAST diagram tells us which trigonometric ratio is positive in that quadrant.

We are dealing with the tangent ratio, and we know that the ratio has a positive value. The A in Quadrant I tells us that all three ratios are positive in that quadrant. And the T in Quadrant III means that the tangent ratio is positive in that quadrant. So Points $W$ and $Y$ are the points which correspond to the equation in the question.

The correct points are $W$ and $Y$ .

Submit your answer as:
In Question 1 you found the two points which correspond to this equation:

$\begin{array}{c}\tan \theta = \dfrac{5}{12}\end{array}$

There are two values for $\theta$ connected to those points. From the list of choices below, select the two values closest to $\theta$ (the choices in the list have been rounded).

TIP: Use the answer to Question 1.

Answer: The values for $\theta$ are closest to: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to pick the angles which match the correct points in Question 1.

STEP: Use the answer to Question 1
[−1 point ⇒ 0 / 1 points left]

For this question we need to identify the values of $\theta$ which correspond to the equation

$\begin{array}{c}\tan \theta = \dfrac{5}{12}\end{array}$

There are two such values: they match the points we identified in Question 1.

We already know that Points $W$ and $Y$ correspond to the equation. So we need to find the angles which match those points. The choices in the list include these four angles: $23° ; 157° ; 203° ; 337°$ . Two of these choices agree with the positions of Points $W$ and $Y$ . (Remember that the angles always start on the positive $x$ -axis, as shown below.)

We do not know the exact values of these two angles. But we know that one of them points into Quadrant I and the other points into Quadrant III. That means the smaller angle must be acute (between $\text{0}\text{°}$ and $\text{90}\text{°}$ ) and the larger angle must be between $\text{180}\text{°}$ and $\text{270}\text{°}$ . Based on the choices in the list, the values for $\theta$ must be $23°$ and $203°$ .

The two values for $\theta$ are $23°$ and $203°$ .

Submit your answer as:

ID is: 3653 Seed is: 4103

Trigonometric ratios on the Cartesian plane: positives & negatives

Consider the following equation about an angle $\theta$ :

$\begin{array}{c}\sin \theta = \dfrac{12}{13}\end{array}$

The diagram below shows four points on the Cartesian plane. They are labelled $W, X, Y$ , and $Z$ . The diagram is not drawn to scale.

Two of the four points above correspond to the equation $\sin \theta = \dfrac{12}{13}$ . Which two?

Answer:
The points corresponding to $\sin \theta = \dfrac{12}{13}$ are Points .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to compare the sign of the ratio to the CAST diagram.

STEP: Use the CAST diagram and the sign of the ratio given
[−1 point ⇒ 0 / 1 points left]

On the Cartesian plane, there is a connection between the sign of the trigonometric ratios and which quadrant the angle points into. For each of the ratios, there are two quadrants where the ratio is positive and two quadrants where the ratio is negative. The CAST diagram shows us these quadrants. So we can use the CAST diagram to answer this question.

In this case we know that:

$\begin{aligned} \sin \theta = & \underbrace{\dfrac{12}{13}} \\ & \hspace{-0.4cm}\text{positive} \end{aligned}$

This equation tells us that the angle $\theta$ must point into one of the two quadrants where the sine ratio is positive. Let's add the CAST diagram letters onto the diagram given in the question to find those two quadrants. Each letter of the CAST diagram tells us which trigonometric ratio is positive in that quadrant.

We are dealing with the sine ratio, and we know that the ratio has a positive value. The A in Quadrant I tells us that all three ratios are positive in that quadrant. And the S in Quadrant II means that the sine ratio is positive in that quadrant. So Points $W$ and $X$ are the points which correspond to the equation in the question.

The correct points are $W$ and $X$ .

Submit your answer as:
In Question 1 you found the two points which correspond to this equation:

$\begin{array}{c}\sin \theta = \dfrac{12}{13}\end{array}$

There are two values for $\theta$ connected to those points. From the list of choices below, select the two values closest to $\theta$ (the choices in the list have been rounded).

TIP: Use the answer to Question 1.

Answer: The values for $\theta$ are closest to: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to pick the angles which match the correct points in Question 1.

STEP: Use the answer to Question 1
[−1 point ⇒ 0 / 1 points left]

For this question we need to identify the values of $\theta$ which correspond to the equation

$\begin{array}{c}\sin \theta = \dfrac{12}{13}\end{array}$

There are two such values: they match the points we identified in Question 1.

We already know that Points $W$ and $X$ correspond to the equation. So we need to find the angles which match those points. The choices in the list include these four angles: $67° ; 113° ; 247° ; 293°$ . Two of these choices agree with the positions of Points $W$ and $X$ . (Remember that the angles always start on the positive $x$ -axis, as shown below.)

We do not know the exact values of these two angles. But we know that one of them points into Quadrant I and the other points into Quadrant II. That means the smaller angle must be acute (between $\text{0}\text{°}$ and $\text{90}\text{°}$ ) and the larger angle must be obtuse (between $\text{90}\text{°}$ and $\text{180}\text{°}$ ). Based on the choices in the list, the values for $\theta$ must be $67°$ and $113°$ .

The two values for $\theta$ are $67°$ and $113°$ .

Submit your answer as:

ID is: 3663 Seed is: 1659

Finding a ratio from a point

The figure below shows Point $K$ at $(x; -5)$ . It is a distance $\text{13}$ from the origin, as labelled. The angle between the positive $x$ -axis and Point $K$ is $B$ . This diagram is not drawn to scale.

Answer the two question which follow about Point $K$ and angle $B$ .

Determine the $x$ -coordinate of Point $K$ .

Answer: The $x$ -coordinate is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Draw a right-angled reference triangle on the figure. Then label the side or sides of the triangle that you know. You can then use the theorem of Pythagoras to find the missing side of the triangle.

STEP: Draw a right-angled triangle onto the given figure
[−1 point ⇒ 2 / 3 points left]

We can find the missing coordinate because the information given lets us draw a right-angled triangle. We already know the hypotenuse of the triangle (it is the radius). We can label another side of the triangle because we know the $y$ -coordinate of Point $K$ .

Notice that the side of the triangle is labelled with a positive number, even though the $y$ -coordinate is negative. See the note below for an explanation about why these signs are not always the same.

STEP: Use the triangle to calculate the missing side
[−1 point ⇒ 1 / 3 points left]

Remember that we want to find the $x$ -coordinate of Point $K$ . We can find it if we know the length of the missing side of the triangle. And we can find that using the theorem of Pythagoras.

$\begin{aligned} a^{2} + b^{2} & = c^{2} \\ x^{2} + (-5)^{2} & = (13)^{2} \\ x^{2} + 25 & = 169 \\ x^{2} & = 144 \\ x & = \pm \sqrt{144} \\ x & = \pm 12 \end{aligned}$

This is the length of the third side of the triangle. It must be positive (it is a length) so the length of the side is $\text{12}$ units.

STEP: Check the sign and adjust it if necessary
[−1 point ⇒ 0 / 3 points left]

The result above tells us that Point $K$ is $\text{12}$ units away from the $y$ -axis. So the missing coordinate must be either $\text{12}$ or $-\text{12}$ . It depends on the position of the point! From the diagram we can see that Point $K$ is in Quadrant IV, so the $x$ -coordinate has to be positive.

NOTE: Sometimes the missing coordinate is negative even though the length of the side of the triangle is positive. Why is this? The answer is that the sides of the triangle are always positive because they are lengths. But for points on the Cartesian plane, we use signs to indicate the position of points: $\text{3}$ spaces right is not the same as $\text{3}$ spaces left. It is crucial to be aware that these signs can be different.

The $x$ -coordinate of Point $K$ is $x = 12$ .

Submit your answer as:
Now determine the value of $\cos (B)$ .

INSTRUCTION: Give your answer as a simplified fraction.

Answer: $\cos (B) =$ .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the coordinates of the point and the radius, whichever are necessary.

STEP: Use the definition of the cosine ratio on the Cartesian plane
[−2 points ⇒ 0 / 2 points left]

The definition of the cosine ratio on the Cartesian plane is:

$\begin{array}{c} \cos (\theta) = \dfrac{x}{r} \end{array}$

Point $K$ corresponds to angle $B$ , so all we need to do is substitute in the correct values from the point. In this case that means using the $x$ -coordinate and the radius.

$\begin{aligned} \cos (\theta) & = \dfrac{x}{r} \\ \cos (B) & = \frac{12}{13} \end{aligned}$

TIP: Compare the sign of your answer to the CAST diagram. Point $K$ is in Quadrant IV. And the CAST diagram tells us that the cosine ratio in Quadrant IV is always positive. Make sure the sign of your answer agrees with the CAST diagram.

The value of $\cos (B)$ is $\dfrac{12}{13}$ .

Submit your answer as:

ID is: 3663 Seed is: 1740

Finding a ratio from a point

The figure below shows Point $M$ at $(x; 8)$ . It is a distance $\text{10}$ from the origin, as labelled. The angle between the positive $x$ -axis and Point $M$ is $B$ . This diagram is not drawn to scale.

Answer the two question which follow about Point $M$ and angle $B$ .

Compute the $x$ -coordinate of Point $M$ .

Answer: The $x$ -coordinate is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Draw a right-angled reference triangle on the figure. Then label the side or sides of the triangle that you know. You can then use the theorem of Pythagoras to find the missing side of the triangle.

STEP: Draw a right-angled triangle onto the given figure
[−1 point ⇒ 2 / 3 points left]

We can find the missing coordinate because the information given lets us draw a right-angled triangle. We already know the hypotenuse of the triangle (it is the radius). We can label another side of the triangle because we know the $y$ -coordinate of Point $M$ .

STEP: Use the triangle to calculate the missing side
[−1 point ⇒ 1 / 3 points left]

Remember that we want to find the $x$ -coordinate of Point $M$ . We can find it if we know the length of the missing side of the triangle. And we can find that using the theorem of Pythagoras.

$\begin{aligned} a^{2} + b^{2} & = c^{2} \\ x^{2} + (8)^{2} & = (10)^{2} \\ x^{2} + 64 & = 100 \\ x^{2} & = 36 \\ x & = \pm \sqrt{36} \\ x & = \pm 6 \end{aligned}$

This is the length of the third side of the triangle. It must be positive (it is a length) so the length of the side is $\text{6}$ units.

STEP: Check the sign and adjust it if necessary
[−1 point ⇒ 0 / 3 points left]

The result above tells us that Point $M$ is $\text{6}$ units away from the $y$ -axis. So the missing coordinate must be either $\text{6}$ or $-\text{6}$ . It depends on the position of the point! From the diagram we can see that Point $M$ is in Quadrant I, so the $x$ -coordinate has to be positive.

NOTE: Sometimes the missing coordinate is negative even though the length of the side of the triangle is positive. Why is this? The answer is that the sides of the triangle are always positive because they are lengths. But for points on the Cartesian plane, we use signs to indicate the position of points: $\text{3}$ spaces right is not the same as $\text{3}$ spaces left. It is crucial to be aware that these signs can be different.

The $x$ -coordinate of Point $M$ is $x = 6$ .

Submit your answer as:
Now determine the value of $\tan (B)$ .

INSTRUCTION: Give your answer as a simplified fraction.

Answer: $\tan (B) =$ .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the coordinates of the point and the radius, whichever are necessary.

STEP: Use the definition of the tangent ratio on the Cartesian plane
[−2 points ⇒ 0 / 2 points left]

The definition of the tangent ratio on the Cartesian plane is:

$\begin{array}{c} \tan (\theta) = \dfrac{y}{x} \end{array}$

Point $M$ corresponds to angle $B$ , so all we need to do is substitute in the correct values from the point. In this case that means using both $x$ and $y$ from Point $P$ .

$\begin{aligned} \tan (\theta) & = \dfrac{y}{x} \\ \tan (B) & = \frac{8}{6} \\ & = \dfrac{4}{3} \end{aligned}$

TIP: Compare the sign of your answer to the CAST diagram. Point $M$ is in Quadrant I. And the CAST diagram tells us that the tangent ratio in Quadrant I is always positive. Make sure the sign of your answer agrees with the CAST diagram.

The value of $\tan (B)$ is $\dfrac{4}{3}$ .

Submit your answer as:

ID is: 3663 Seed is: 5587

Finding a ratio from a point

The figure below shows Point $A$ at $(x; 12)$ . It is a distance $\text{20}$ from the origin, as labelled. The angle between the positive $x$ -axis and Point $A$ is $B$ . This diagram is not drawn to scale.

Answer the two question which follow about Point $A$ and angle $B$ .

Find the $x$ -coordinate of Point $A$ .

Answer: The $x$ -coordinate is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Draw a right-angled reference triangle on the figure. Then label the side or sides of the triangle that you know. You can then use the theorem of Pythagoras to find the missing side of the triangle.

STEP: Draw a right-angled triangle onto the given figure
[−1 point ⇒ 2 / 3 points left]

We can find the missing coordinate because the information given lets us draw a right-angled triangle. We already know the hypotenuse of the triangle (it is the radius). We can label another side of the triangle because we know the $y$ -coordinate of Point $A$ .

STEP: Use the triangle to calculate the missing side
[−1 point ⇒ 1 / 3 points left]

Remember that we want to find the $x$ -coordinate of Point $A$ . We can find it if we know the length of the missing side of the triangle. And we can find that using the theorem of Pythagoras.

$\begin{aligned} a^{2} + b^{2} & = c^{2} \\ x^{2} + (12)^{2} & = (20)^{2} \\ x^{2} + 144 & = 400 \\ x^{2} & = 256 \\ x & = \pm \sqrt{256} \\ x & = \pm 16 \end{aligned}$

This is the length of the third side of the triangle. It must be positive (it is a length) so the length of the side is $\text{16}$ units.

STEP: Check the sign and adjust it if necessary
[−1 point ⇒ 0 / 3 points left]

The result above tells us that Point $A$ is $\text{16}$ units away from the $y$ -axis. So the missing coordinate must be either $\text{16}$ or $-\text{16}$ . It depends on the position of the point! From the diagram we can see that Point $A$ is in Quadrant I, so the $x$ -coordinate has to be positive.

NOTE: Sometimes the missing coordinate is negative even though the length of the side of the triangle is positive. Why is this? The answer is that the sides of the triangle are always positive because they are lengths. But for points on the Cartesian plane, we use signs to indicate the position of points: $\text{3}$ spaces right is not the same as $\text{3}$ spaces left. It is crucial to be aware that these signs can be different.

The $x$ -coordinate of Point $A$ is $x = 16$ .

Submit your answer as:
Now determine the value of $\cos (B)$ .

INSTRUCTION: Give your answer as a simplified fraction.

Answer: $\cos (B) =$ .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the coordinates of the point and the radius, whichever are necessary.

STEP: Use the definition of the cosine ratio on the Cartesian plane
[−2 points ⇒ 0 / 2 points left]

The definition of the cosine ratio on the Cartesian plane is:

$\begin{array}{c} \cos (\theta) = \dfrac{x}{r} \end{array}$

Point $A$ corresponds to angle $B$ , so all we need to do is substitute in the correct values from the point. In this case that means using the $x$ -coordinate and the radius.

$\begin{aligned} \cos (\theta) & = \dfrac{x}{r} \\ \cos (B) & = \frac{16}{20} \\ & = \dfrac{4}{5} \end{aligned}$

TIP: Compare the sign of your answer to the CAST diagram. Point $A$ is in Quadrant I. And the CAST diagram tells us that the cosine ratio in Quadrant I is always positive. Make sure the sign of your answer agrees with the CAST diagram.

The value of $\cos (B)$ is $\dfrac{4}{5}$ .

Submit your answer as:

ID is: 3666 Seed is: 8563

Finding one ratio from another

Given:

\begin{matrix} \sin (B) = - \frac{13}{14} \end{matrix}

$\begin{array}{c} \sin (B) = -\dfrac{13}{14} \end{array}$

Determine the value of $\cos (B)$ if $B$ is greater than $270°$ and less than $360°$ .

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a square root, type sqrt( ).

Answer:

\cos (B) =

$\cos (B) =$ .

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by figuring out which quadrant the angle $B$ is in and draw a sketch to represent that angle.

STEP: Sketch the angle and a reference triangle
[−2 points ⇒ 2 / 4 points left]

The first thing to do is figure out which quadrant the angle $B$ points into. Then we can draw a sketch corresponding to that angle, including a reference triangle.

Based on the fact that the angle is greater than $270°$ and less than $360°$ , and the fact that $\sin (B)$ is negative, the angle must be in Quadrant IV. So we can draw and label a diagram as follows:

The labels for the coordinates and the sides of the triangle are based on the fact that the sine ratio is the ratio of $y$ to $r$ . While the value of $\sin (B)$ is negative, the labels on the triangle are positive because lengths are always positive.

NOTE: The figure above is a sketch: the triangle is probably not actually the size shown above. That is not a problem, but it means that we cannot trust the appearance of the figure. Instead, we can only trust the labels.

STEP: Find the value of

\cos (B)

$\cos (B)$
[−2 points ⇒ 0 / 4 points left]

We want to find $\cos (B)$ , which means we want the ratio $\dfrac{x}{r}$ . So we need the value of $x$ . We can find this using the theorem of Pythagoras with the reference triangle:

\begin{aligned} a^{2} + b^{2} & = c^{2} \\ x^{2} + (13)^{2} & = (14)^{2} \\ x^{2} & = 196 - 169 \\ x^{2} & = 27 \\ x^{2} & = 9 \cdot 3 \\ \sqrt{x^{2}} & = \pm \sqrt{9 \cdot 3} \\ x & = \pm 3 \sqrt{3} \end{aligned}

$\begin{aligned} a^{2} + b^{2} & = c^{2} \\ x^{2} + (13)^{2} & = (14)^{2} \\ x^{2} & = 196 - 169 \\ x^{2} & = 27 \\ x^{2} & = 9 \cdot 3 \\ \sqrt{ x^{2} } & = \pm \sqrt{ 9 \cdot 3 } \\ x & = \pm 3 \sqrt{3} \end{aligned}$

This calculation tells us two things: the third side of the triangle is $3 \sqrt{3}$ units long, and the $x$ -coordinate of the point in our diagram is $3 \sqrt{3}$ . Now we have enough information to evaluate the cosine ratio for angle $B$ .

\begin{aligned} \cos (B) & = \frac{x}{r} \\ = \frac{3 \sqrt{3}}{14} \end{aligned}

$\begin{aligned} \cos (B) & = \dfrac{x}{r} \\ & = \dfrac{3 \sqrt{3}}{14} \end{aligned}$

TIP: Remember to check the sign of your answer against the CAST diagram. The angle is in Quadrant IV and we expect the cosine ratio to be positive in this quadrant. So all is groovy!

The value of $\cos (B)$ is $\dfrac{3 \sqrt{3}}{14}$ .

Submit your answer as:

ID is: 3666 Seed is: 7598

Finding one ratio from another

Given:

\begin{matrix} \cos (C) = \frac{7}{13} \end{matrix}

$\begin{array}{c} \cos (C) = \dfrac{7}{13} \end{array}$

Determine the value of $\sin (C)$ if $C$ is greater than $180°$ and less than $360^{\circ$.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a square root, type sqrt( ).

Answer:

\sin (C) =

$\sin (C) =$ .

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by figuring out which quadrant the angle $C$ is in and draw a sketch to represent that angle.

STEP: Sketch the angle and a reference triangle
[−2 points ⇒ 2 / 4 points left]

The first thing to do is figure out which quadrant the angle $C$ points into. Then we can draw a sketch corresponding to that angle, including a reference triangle.

Based on the fact that the angle is greater than $180°$ and less than $360^{\circ$, and the fact that $\cos (C)$ is positive, the angle must be in Quadrant IV. So we can draw and label a diagram as follows:

The labels for the coordinates and the sides of the triangle are based on the fact that the cosine ratio is the ratio of $x$ to $r$ .

STEP: Find the value of

\sin (C)

$\sin (C)$
[−2 points ⇒ 0 / 4 points left]

We want to find $\sin (C)$ , which means we want the ratio $\dfrac{y}{r}$ . So we need the value of $y$ . We can find this using the theorem of Pythagoras with the reference triangle:

\begin{aligned} a^{2} + b^{2} & = c^{2} \\ y^{2} + (7)^{2} & = (13)^{2} \\ y^{2} & = 169 - 49 \\ y^{2} & = 120 \\ y^{2} & = 4 \cdot 30 \\ \sqrt{y^{2}} & = \pm \sqrt{4 \cdot 30} \\ y & = \pm 2 \sqrt{30} \end{aligned}

$\begin{aligned} a^{2} + b^{2} & = c^{2} \\ y^{2} + (7)^{2} & = (13)^{2} \\ y^{2} & = 169 - 49 \\ y^{2} & = 120 \\ y^{2} & = 4 \cdot 30 \\ \sqrt{ y^{2} } & = \pm \sqrt{ 4 \cdot 30 } \\ y & = \pm 2 \sqrt{30} \end{aligned}$

This calculation tells us two things: the third side of the triangle is $2 \sqrt{30}$ units long, and the $y$ -coordinate of the point in our diagram is $- 2 \sqrt{30}$ . Now we have enough information to evaluate the sine ratio for angle $C$ .

\begin{aligned} \sin (C) & = \frac{y}{r} \\ = \frac{- 2 \sqrt{30}}{13} \\ = - \frac{2 \sqrt{30}}{13} \end{aligned}

$\begin{aligned} \sin (C) & = \dfrac{y}{r} \\ & = \dfrac{- 2 \sqrt{30}}{13} \\ & = -\dfrac{2 \sqrt{30}}{13} \end{aligned}$

TIP: Remember to check the sign of your answer against the CAST diagram. The angle is in Quadrant IV and we expect the sine ratio to be negative in this quadrant. So all is groovy!

The value of $\sin (C)$ is $-\dfrac{2 \sqrt{30}}{13}$ .

Submit your answer as:

ID is: 3666 Seed is: 1837

Finding one ratio from another

Given:

\begin{matrix} \cos (B) = \frac{7}{9} \end{matrix}

$\begin{array}{c} \cos (B) = \dfrac{7}{9} \end{array}$

Determine the value of $\tan (B)$ if $B$ is acute.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a square root, type sqrt( ).

Answer:

\tan (B) =

$\tan (B) =$ .

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by figuring out which quadrant the angle $B$ is in and draw a sketch to represent that angle.

STEP: Sketch the angle and a reference triangle
[−2 points ⇒ 2 / 4 points left]

The first thing to do is figure out which quadrant the angle $B$ points into. Then we can draw a sketch corresponding to that angle, including a reference triangle.

Based on the fact that the angle is acute, the angle must be in Quadrant I. This agrees with the fact that $\cos (B)$ is positive. So we can draw and label a diagram as follows:

The labels for the coordinates and the sides of the triangle are based on the fact that the cosine ratio is the ratio of $x$ to $r$ .

STEP: Find the value of

\tan (B)

$\tan (B)$
[−2 points ⇒ 0 / 4 points left]

We want to find $\tan (B)$ , which means we want the ratio $\dfrac{y}{x}$ . So we need the value of $y$ . We can find this using the theorem of Pythagoras with the reference triangle:

\begin{aligned} a^{2} + b^{2} & = c^{2} \\ y^{2} + (7)^{2} & = (9)^{2} \\ y^{2} & = 81 - 49 \\ y^{2} & = 32 \\ y^{2} & = 16 \cdot 2 \\ \sqrt{y^{2}} & = \pm \sqrt{16 \cdot 2} \\ y & = \pm 4 \sqrt{2} \end{aligned}

$\begin{aligned} a^{2} + b^{2} & = c^{2} \\ y^{2} + (7)^{2} & = (9)^{2} \\ y^{2} & = 81 - 49 \\ y^{2} & = 32 \\ y^{2} & = 16 \cdot 2 \\ \sqrt{ y^{2} } & = \pm \sqrt{ 16 \cdot 2 } \\ y & = \pm 4 \sqrt{2} \end{aligned}$

This calculation tells us two things: the third side of the triangle is $4 \sqrt{2}$ units long, and the $y$ -coordinate of the point in our diagram is $4 \sqrt{2}$ . Now we have enough information to evaluate the tangent ratio for angle $B$ .

\begin{aligned} \tan (B) & = \frac{y}{x} \\ = \frac{4 \sqrt{2}}{7} \end{aligned}

$\begin{aligned} \tan (B) & = \dfrac{y}{x} \\ & = \dfrac{4 \sqrt{2}}{7} \end{aligned}$

TIP: Remember to check the sign of your answer against the CAST diagram. The angle is in Quadrant I and we expect the tangent ratio to be positive in this quadrant. So all is groovy!

The value of $\tan (B)$ is $\dfrac{4 \sqrt{2}}{7}$ .

Submit your answer as:

ID is: 3601 Seed is: 7949

The purpose of the CAST diagram

The diagram below is often called the CAST diagram.

What does the CAST diagram tell us?

Answer:

The CAST diagram tells us .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find a section on the CAST diagram in the Everything Maths textbook.

STEP: Identify the purpose of the CAST diagram
[−1 point ⇒ 0 / 1 points left]

This question is about the CAST diagram. You will find it useful to memorise the CAST diagram! Each letter in CAST refers to trigonometric ratios:

C - cosine
A - all (sine, cosine, and tangent)
S - sine
T - tangent

For each quadrant, the CAST diagram shows us which trigonometric ratios are positive in that quadrant. For example, let's focus on Quadrant II, where we find the letter S:

The S is always in Quadrant II. The sine ratio is positive for any angle in Quadrant II. At the same time, it tells us that the cosine and tangent ratios are negative in that quadrant.

Here are some useful facts about the CAST diagram:

In Quadrant I all the trigonometric ratios are positive. This means all three ratios are positive for angles between $0°$ and $90°$ .
Each trigonometric ratio is positive in $\text{2}$ of the quadrants and negative in $\text{2}$ of the quadrants. For example, the sine ratio is positive in Quadrants I and II, while it is negative in Quadrants III and IV.

The CAST diagram tells us when the trigonometric ratios are positive or negative.

Submit your answer as:

ID is: 3601 Seed is: 3612

The purpose of the CAST diagram

The diagram below is often called the CAST diagram.

What does the CAST diagram tell us?

Answer:

The CAST diagram tells us .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find a section on the CAST diagram in the Everything Maths textbook.

STEP: Identify the purpose of the CAST diagram
[−1 point ⇒ 0 / 1 points left]

This question is about the CAST diagram. You will find it useful to memorise the CAST diagram! Each letter in CAST refers to trigonometric ratios:

C - cosine
A - all (sine, cosine, and tangent)
S - sine
T - tangent

For each quadrant, the CAST diagram shows us which trigonometric ratios are positive in that quadrant. For example, let's focus on Quadrant III, where we find the letter T:

The T is always in Quadrant III. The tangent ratio is positive for any angle in Quadrant III. At the same time, it tells us that the sine and cosine ratios are negative in that quadrant.

Here are some useful facts about the CAST diagram:

In Quadrant I all the trigonometric ratios are positive. This means all three ratios are positive for angles between $0°$ and $90°$ .
Each trigonometric ratio is positive in $\text{2}$ of the quadrants and negative in $\text{2}$ of the quadrants. For example, the tangent ratio is positive in Quadrants I and III, while it is negative in Quadrants II and IV.

The CAST diagram tells us when the trigonometric ratios are positive or negative.

Submit your answer as:

ID is: 3601 Seed is: 4968

The purpose of the CAST diagram

The diagram below is often called the CAST diagram.

What does the CAST diagram tell us?

Answer:

The CAST diagram tells us .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find a section on the CAST diagram in the Everything Maths textbook.

STEP: Identify the purpose of the CAST diagram
[−1 point ⇒ 0 / 1 points left]

This question is about the CAST diagram. You will find it useful to memorise the CAST diagram! Each letter in CAST refers to trigonometric ratios:

C - cosine
A - all (sine, cosine, and tangent)
S - sine
T - tangent

For each quadrant, the CAST diagram shows us which trigonometric ratios are positive in that quadrant. For example, let's focus on Quadrant IV, where we find the letter C:

The C is always in Quadrant IV. The cosine ratio is positive for any angle in Quadrant IV. At the same time, it tells us that the sine and tangent ratios are negative in that quadrant.

Here are some useful facts about the CAST diagram:

In Quadrant I all the trigonometric ratios are positive. This means all three ratios are positive for angles between $0°$ and $90°$ .
Each trigonometric ratio is positive in $\text{2}$ of the quadrants and negative in $\text{2}$ of the quadrants. For example, the tangent ratio is positive in Quadrants I and III, while it is negative in Quadrants II and IV.

The CAST diagram tells us when the trigonometric ratios are positive or negative.

Submit your answer as:

ID is: 3665 Seed is: 3004

Trigonometric ratios and the radius

The figure below shows a circle centred at the origin. Point $F$ is on the circle at $(12; 5)$ . The angle between the positive $x$ -axis and Point $F$ is $f$ . A radius of the circle is shown from the origin to Point $F$ .

Answer the two questions which follow about Point $F$ and $f$ .

Determine the length of the radius, $r$ .

Answer: The length of the radius is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by drawing a right-angled reference triangle on the figure. This triangle should include Point $F$ and the radius shown.

STEP: Draw a right-angled triangle on the figure
[−1 point ⇒ 1 / 2 points left]

We need to find the radius of the circle. We can do this by using the coordinates of Point $F$ . This first thing we need to do is draw a right-angled triangle on the figure. This triangle should connect Point $F$ , the origin, and the $x$ -axis, as shown below.

Here is the key step: we can label the lengths of the sides of the triangle based on the coordinates of Point $F$ . The $x$ -coordinate of Point $F$ is $\text{12}$ : that means $F$ is $\text{12}$ units away from the $y$ -axis. Similarly, the $y$ -coordinate of $F$ is $\text{5}$ , which means that $F$ is $\text{5}$ units away from the $x$ -axis.

TIP: You can count the grid marks on the picture to verify that the lengths of the triangle's legs are correct!

STEP: Use the triangle to calculate the radius
[−1 point ⇒ 0 / 2 points left]

The radius that we want is the hypotenuse of the triangle. So we can find the radius using the theorem of Pythagoras.

$\begin{aligned} a^2 + b^2 & = c^2 \\ (12)^2 + (5)^2 & = r^2 \\ 169 & = r^2 \\ \pm \sqrt{169} & = \sqrt{r^2} \\ \pm 13 & = r \end{aligned}$

We get a radius of $\pm 13$ . But it must be positive because it is a length.

The length of the radius of the circle is $\text{13}$ .

Submit your answer as:
Hence determine the value of $\sin (f)$ .

Answer: The value of $\sin (f)$ is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the definition of the sine ratio on the Cartesian plane. Then use the coordinates of the point and the radius to calculate the answer.

STEP: Use the definition of the sine ratio on the Cartesian plane
[−2 points ⇒ 0 / 2 points left]

We need to find the value of $\sin (f)$ . The trigonometric functions are defined on the Cartesian plane in terms of the coordinates $(x; y)$ and the radius $r$ . And for $f$ , we know all three of those numbers from Question 1! In this case we want to know $\sin (f)$ , so we need the definition of the sine ratio on the Cartesian plane:

$\begin{array}{c} \sin (\theta) = \dfrac{y}{r} \end{array}$

Point $F$ corresponds to $f$ , so we need to substitute in the correct values from the point. In this case that means using the $y$ -coordinate and the radius.

$\begin{aligned} y & = 5 \\ r & = 13 \\[0.2cm] \sin (f) & = \frac{5}{13} \end{aligned}$

TIP: Compare the sign of your answer to the CAST diagram. Point $F$ is in Quadrant I. The CAST diagram tells us that the sine ratio in Quadrant I is always positive. Make sure the sign of your answer agrees with the CAST diagram.

The value of $\sin (f)$ is $\dfrac{\text{5}}{\text{13}}$ .

Submit your answer as:

ID is: 3665 Seed is: 5094

Trigonometric ratios and the radius

The figure below shows a circle centred at the origin. Point $E$ is on the circle at $(6; 8)$ . The angle between the positive $x$ -axis and Point $E$ is $e$ . A radius of the circle is shown from the origin to Point $E$ .

Answer the two questions which follow about Point $E$ and $e$ .

Determine the length of the radius, $r$ .

Answer: The length of the radius is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by drawing a right-angled reference triangle on the figure. This triangle should include Point $E$ and the radius shown.

STEP: Draw a right-angled triangle on the figure
[−1 point ⇒ 1 / 2 points left]

We need to find the radius of the circle. We can do this by using the coordinates of Point $E$ . This first thing we need to do is draw a right-angled triangle on the figure. This triangle should connect Point $E$ , the origin, and the $x$ -axis, as shown below.

Here is the key step: we can label the lengths of the sides of the triangle based on the coordinates of Point $E$ . The $x$ -coordinate of Point $E$ is $\text{6}$ : that means $E$ is $\text{6}$ units away from the $y$ -axis. Similarly, the $y$ -coordinate of $E$ is $\text{8}$ , which means that $E$ is $\text{8}$ units away from the $x$ -axis.

TIP: You can count the grid marks on the picture to verify that the lengths of the triangle's legs are correct!

STEP: Use the triangle to calculate the radius
[−1 point ⇒ 0 / 2 points left]

The radius that we want is the hypotenuse of the triangle. So we can find the radius using the theorem of Pythagoras.

$\begin{aligned} a^2 + b^2 & = c^2 \\ (6)^2 + (8)^2 & = r^2 \\ 100 & = r^2 \\ \pm \sqrt{100} & = \sqrt{r^2} \\ \pm 10 & = r \end{aligned}$

We get a radius of $\pm 10$ . But it must be positive because it is a length.

The length of the radius of the circle is $\text{10}$ .

Submit your answer as:
Hence determine the value of $\cos (e)$ .

Answer: The value of $\cos (e)$ is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the definition of the cosine ratio on the Cartesian plane. Then use the coordinates of the point and the radius to calculate the answer.

STEP: Use the definition of the cosine ratio on the Cartesian plane
[−2 points ⇒ 0 / 2 points left]

We need to find the value of $\cos (e)$ . The trigonometric functions are defined on the Cartesian plane in terms of the coordinates $(x; y)$ and the radius $r$ . And for $e$ , we know all three of those numbers from Question 1! In this case we want to know $\cos (e)$ , so we need the definition of the cosine ratio on the Cartesian plane:

$\begin{array}{c} \cos (\theta) = \dfrac{x}{r} \end{array}$

Point $E$ corresponds to $e$ , so we need to substitute in the correct values from the point. In this case that means using the $x$ -coordinate and the radius.

$\begin{aligned} x & = 6 \\ r & = 10 \\[0.2cm] \cos (e) & = \frac{6}{10} \\ & =\frac{3}{5} \end{aligned}$

TIP: Compare the sign of your answer to the CAST diagram. Point $E$ is in Quadrant I. The CAST diagram tells us that the cosine ratio in Quadrant I is always positive. Make sure the sign of your answer agrees with the CAST diagram.

The value of $\cos (e)$ is $\dfrac{\text{3}}{\text{5}}$ .

Submit your answer as:

ID is: 3665 Seed is: 6340

Trigonometric ratios and the radius

The figure below shows a circle centred at the origin. Point $F$ is on the circle at $(-5; -12)$ . The angle between the positive $x$ -axis and Point $F$ is $f$ . A radius of the circle is shown from the origin to Point $F$ .

Answer the two questions which follow about Point $F$ and $f$ .

Determine the length of the radius, $r$ .

Answer: The length of the radius is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by drawing a right-angled reference triangle on the figure. This triangle should include Point $F$ and the radius shown.

STEP: Draw a right-angled triangle on the figure
[−1 point ⇒ 1 / 2 points left]

We need to find the radius of the circle. We can do this by using the coordinates of Point $F$ . This first thing we need to do is draw a right-angled triangle on the figure. This triangle should connect Point $F$ , the origin, and the $x$ -axis, as shown below.

Here is the key step: we can label the lengths of the sides of the triangle based on the coordinates of Point $F$ . The $x$ -coordinate of Point $F$ is $-\text{5}$ : that means $F$ is $\text{5}$ units away from the $y$ -axis. Similarly, the $y$ -coordinate of $F$ is $-\text{12}$ , which means that $F$ is $\text{12}$ units away from the $x$ -axis.

Notice that the sides of the triangle are labelled with positive numbers even though the $x$ - and $y$ -coordinates are negative.

TIP: You can count the grid marks on the picture to verify that the lengths of the triangle's legs are correct!

STEP: Use the triangle to calculate the radius
[−1 point ⇒ 0 / 2 points left]

The radius that we want is the hypotenuse of the triangle. So we can find the radius using the theorem of Pythagoras.

$\begin{aligned} a^2 + b^2 & = c^2 \\ (5)^2 + (12)^2 & = r^2 \\ 169 & = r^2 \\ \pm \sqrt{169} & = \sqrt{r^2} \\ \pm 13 & = r \end{aligned}$

We get a radius of $\pm 13$ . But it must be positive because it is a length.

The length of the radius of the circle is $\text{13}$ .

Submit your answer as:
Hence determine the value of $\cos (f)$ .

Answer: The value of $\cos (f)$ is .
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the definition of the cosine ratio on the Cartesian plane. Then use the coordinates of the point and the radius to calculate the answer.

STEP: Use the definition of the cosine ratio on the Cartesian plane
[−2 points ⇒ 0 / 2 points left]

We need to find the value of $\cos (f)$ . The trigonometric functions are defined on the Cartesian plane in terms of the coordinates $(x; y)$ and the radius $r$ . And for $f$ , we know all three of those numbers from Question 1! In this case we want to know $\cos (f)$ , so we need the definition of the cosine ratio on the Cartesian plane:

$\begin{array}{c} \cos (\theta) = \dfrac{x}{r} \end{array}$

Point $F$ corresponds to $f$ , so we need to substitute in the correct values from the point. In this case that means using the $x$ -coordinate and the radius.

$\begin{aligned} x & = -5 \\ r & = 13 \\[0.2cm] \cos (f) & = - \frac{5}{13} \end{aligned}$

TIP: Compare the sign of your answer to the CAST diagram. Point $F$ is in Quadrant III. The CAST diagram tells us that the cosine ratio in Quadrant III is always negative. Make sure the sign of your answer agrees with the CAST diagram.

The value of $\cos (f)$ is $-\dfrac{\text{5}}{\text{13}}$ .

Submit your answer as:

ID is: 3664 Seed is: 2027

Finding one ratio from another

Suppose $\cos (\theta) = \dfrac{12}{13}$ , and $0° \leq \theta \leq 180°$ . Without using a calculator, determine the value of $\tan (\theta)$ .

INSTRUCTION: Write your answer as a simplified fraction.

Answer: The value of

\tan (θ)

$\tan (\theta)$ is .

fraction

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by comparing the information given to the CAST diagram to find the quadrant in which $\theta$ sits. Then draw a sketch to represent that angle, including a reference triangle. Use that triangle to find the information you need to find $\tan (\theta)$ .

STEP: Determine which quadrant

θ

$\theta$ must be in
[−1 point ⇒ 4 / 5 points left]

This question is about the expressions $\cos (\theta)$ and $\tan (\theta)$ . We know one and need to find the other. We can do this because $\cos (\theta)$ tells us about two of the three values $x$ , $y$ , and $r$ . We can use those to find the third value, which we need in order to find $\tan (\theta)$ .

The first step is to figure out which quadrant the angle $\theta$ points to. We can do this by combining the CAST diagram with the interval given in the question: $0° \leq \theta \leq 180°$ . We already know that $\cos (\theta)$ is positive because it is equal to $\dfrac{12}{13}$ . The CAST diagram tells us that the cosine ratio is positive in Quadrants I and IV. But since $\theta$ is between $0°$ and $180°$ , it cannot be in Quadrant IV.

NOTE: If we did not know that

0 ° \leq θ \leq 180 °

$0° \leq \theta \leq 180°$ , we could not know if the angle is in Quadrant I or IV.

The angle $\theta$ must point to Quadrant I.

STEP: Draw a reference triangle and label it
[−2 points ⇒ 2 / 5 points left]

Now we can sketch a point in Quadrant I to represent angle $\theta$ .

NOTE: We cannot draw the triangle precisely because we do not know the angle. So a sketch will have to do. The sketch should agree with whatever information we know. For example, we know the point is in Quadrant I so we should put it there. Since the sketch is not precise, we cannot trust the appearance of the figure. Instead, we can only trust the labels.

Now we need to unpack the fact that $\cos (\theta)$ is equal to $\dfrac{12}{13}$ (this is given in the question). This tells us about two sides of the triangle in the picture above.

\begin{aligned} \cos (θ) = \frac{12}{13} & = \frac{x}{r} \\ ↓ \\ \frac{x}{r} & = \frac{12}{13} \\ This means: \\ x & = 12 \\ r & = 13 \end{aligned}

$\begin{aligned} \cos (\theta) = \dfrac{12}{13} & = \dfrac{x}{r} \\ & \downarrow \\ \dfrac{x}{r} & = \dfrac{12}{13} \\ \text{This means:} & \\ x & = 12 \\ r & = 13 \end{aligned}$

Now we know one of the coordinates of Point $P$ . We also know two sides of the triangle. Note that the lengths of the triangle's sides must be positive because they are distances, no matter which quadrant the triangle is in.

STEP: Find the value of

y

$y$
[−1 point ⇒ 1 / 5 points left]

To find the ratio we want, we need the value of $y$ . We can find this using the theorem of Pythagoras for the reference triangle in the figure.

\begin{aligned} a^{2} + b^{2} & = c^{2} \\ y^{2} + (12)^{2} & = (13)^{2} \\ y^{2} & = 169 - 144 \\ y^{2} & = 25 \\ y & = \pm 5 \end{aligned}

$\begin{aligned} a^{2} + b^{2} & = c^{2} \\ y^{2} + (12)^{2} & = (13)^{2} \\ y^{2} & = 169 - 144 \\ y^{2} & = 25 \\ y & = \pm 5 \end{aligned}$

For the third side of the triangle, we must take the positive value (because it is a distance). However, the sign of $y$ depends on which quadrant the point is in. In this case, Point $P$ is in Quadrant I, where the $y$ -values must be positive. So the correct value for the $y$ -coordinate is $\text{5}$ (we throw away the negative answer).

STEP: Use the definition of the tangent ratio on the Cartesian plane
[−1 point ⇒ 0 / 5 points left]

Based on the calculation above we can complete the information about the triangle:

Since we know that Point $P$ corresponds to angle $\theta$ , we can evaluate $\tan (\theta)$ by substituting in the correct values from the point.

\begin{aligned} \tan (θ) & = \frac{y}{x} \\ = \frac{5}{12} \end{aligned}

$\begin{aligned} \tan (\theta) & = \dfrac{y}{x} \\ & = \dfrac{5}{12} \end{aligned}$

TIP: Point

P

$P$ is in Quadrant I, where the CAST diagram tells us that the tangent ratio is always positive. Make sure your answer agrees with that!

The value of $\tan (\theta)$ is $\dfrac{5}{12}$ .

Submit your answer as:

ID is: 3664 Seed is: 582

Finding one ratio from another

Suppose $\cos (\omega) = \dfrac{20}{29}$ , and $180° \leq \omega \leq 360°$ . Without using a calculator, determine the value of $\sin (\omega)$ .

INSTRUCTION: Write your answer as a simplified fraction.

Answer: The value of

\sin (ω)

$\sin (\omega)$ is .

fraction

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by comparing the information given to the CAST diagram to find the quadrant in which $\omega$ sits. Then draw a sketch to represent that angle, including a reference triangle. Use that triangle to find the information you need to find $\sin (\omega)$ .

STEP: Determine which quadrant

ω

$\omega$ must be in
[−1 point ⇒ 4 / 5 points left]

This question is about the expressions $\cos (\omega)$ and $\sin (\omega)$ . We know one and need to find the other. We can do this because $\cos (\omega)$ tells us about two of the three values $x$ , $y$ , and $r$ . We can use those to find the third value, which we need in order to find $\sin (\omega)$ .

The first step is to figure out which quadrant the angle $\omega$ points to. We can do this by combining the CAST diagram with the interval given in the question: $180° \leq \omega \leq 360°$ . We already know that $\cos (\omega)$ is positive because it is equal to $\dfrac{20}{29}$ . The CAST diagram tells us that the cosine ratio is positive in Quadrants I and IV. But since $\omega$ is between $180°$ and $360°$ , it cannot be in Quadrant I.

NOTE: If we did not know that

180 ° \leq ω \leq 360 °

$180° \leq \omega \leq 360°$ , we could not know if the angle is in Quadrant I or IV.

The angle $\omega$ must point to Quadrant IV.

STEP: Draw a reference triangle and label it
[−2 points ⇒ 2 / 5 points left]

Now we can sketch a point in Quadrant IV to represent angle $\omega$ .

NOTE: We cannot draw the triangle precisely because we do not know the angle. So a sketch will have to do. The sketch should agree with whatever information we know. For example, we know the point is in Quadrant IV so we should put it there. Since the sketch is not precise, we cannot trust the appearance of the figure. Instead, we can only trust the labels.

Now we need to unpack the fact that $\cos (\omega)$ is equal to $\dfrac{20}{29}$ (this is given in the question). This tells us about two sides of the triangle in the picture above.

\begin{aligned} \cos (θ) = \frac{20}{29} & = \frac{x}{r} \\ ↓ \\ \frac{x}{r} & = \frac{20}{29} \\ This means: \\ x & = 20 \\ r & = 29 \end{aligned}

$\begin{aligned} \cos (\theta) = \dfrac{20}{29} & = \dfrac{x}{r} \\ & \downarrow \\ \dfrac{x}{r} & = \dfrac{20}{29} \\ \text{This means:} & \\ x & = 20 \\ r & = 29 \end{aligned}$

STEP: Find the value of

y

$y$
[−1 point ⇒ 1 / 5 points left]

To find the ratio we want, we need the value of $y$ . We can find this using the theorem of Pythagoras for the reference triangle in the figure.

\begin{aligned} a^{2} + b^{2} & = c^{2} \\ y^{2} + (20)^{2} & = (29)^{2} \\ y^{2} & = 841 - 400 \\ y^{2} & = 441 \\ y & = \pm 21 \end{aligned}

$\begin{aligned} a^{2} + b^{2} & = c^{2} \\ y^{2} + (20)^{2} & = (29)^{2} \\ y^{2} & = 841 - 400 \\ y^{2} & = 441 \\ y & = \pm 21 \end{aligned}$

For the third side of the triangle, we must take the positive value (because it is a distance). However, the sign of $y$ depends on which quadrant the point is in. In this case, Point $P$ is in Quadrant IV, where the $y$ -values must be negative. So the correct value for the $y$ -coordinate is $-\text{21}$ (we throw away the positive answer).

STEP: Use the definition of the sine ratio on the Cartesian plane
[−1 point ⇒ 0 / 5 points left]

Based on the calculation above we can complete the information about the triangle:

Since we know that Point $P$ corresponds to angle $\omega$ , we can evaluate $\sin (\omega)$ by substituting in the correct values from the point.

\begin{aligned} \sin (θ) & = \frac{y}{r} \\ = - \frac{21}{29} \end{aligned}

$\begin{aligned} \sin (\theta) & = \dfrac{y}{r} \\ & = - \dfrac{21}{29} \end{aligned}$

TIP: Point

P

$P$ is in Quadrant IV, where the CAST diagram tells us that the sine ratio is always negative. Make sure your answer agrees with that!

The value of $\sin (\omega)$ is $- \dfrac{21}{29}$ .

Submit your answer as:

ID is: 3664 Seed is: 3272

Finding one ratio from another

Suppose $\sin (\alpha) = - \dfrac{15}{17}$ , and $90° \leq \alpha \leq 270°$ . Without using a calculator, determine the value of $\tan (\alpha)$ .

INSTRUCTION: Write your answer as a simplified fraction.

Answer: The value of

\tan (α)

$\tan (\alpha)$ is .

fraction

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by comparing the information given to the CAST diagram to find the quadrant in which $\alpha$ sits. Then draw a sketch to represent that angle, including a reference triangle. Use that triangle to find the information you need to find $\tan (\alpha)$ .

STEP: Determine which quadrant

α

$\alpha$ must be in
[−1 point ⇒ 4 / 5 points left]

This question is about the expressions $\sin (\alpha)$ and $\tan (\alpha)$ . We know one and need to find the other. We can do this because $\sin (\alpha)$ tells us about two of the three values $x$ , $y$ , and $r$ . We can use those to find the third value, which we need in order to find $\tan (\alpha)$ .

The first step is to figure out which quadrant the angle $\alpha$ points to. We can do this by combining the CAST diagram with the interval given in the question: $90° \leq \alpha \leq 270°$ . We already know that $\sin (\alpha)$ is negative because it is equal to $- \dfrac{15}{17}$ . The CAST diagram tells us that the sine ratio is negative in Quadrants III and IV. But since $\alpha$ is between $90°$ and $270°$ , it cannot be in Quadrant IV.

NOTE: If we did not know that

90 ° \leq α \leq 270 °

$90° \leq \alpha \leq 270°$ , we could not know if the angle is in Quadrant III or IV.

The angle $\alpha$ must point to Quadrant III.

STEP: Draw a reference triangle and label it
[−2 points ⇒ 2 / 5 points left]

Now we can sketch a point in Quadrant III to represent angle $\alpha$ .

NOTE: We cannot draw the triangle precisely because we do not know the angle. So a sketch will have to do. The sketch should agree with whatever information we know. For example, we know the point is in Quadrant III so we should put it there. Since the sketch is not precise, we cannot trust the appearance of the figure. Instead, we can only trust the labels.

Now we need to unpack the fact that $\sin (\alpha)$ is equal to $- \dfrac{15}{17}$ (this is given in the question). This tells us about two sides of the triangle in the picture above.

\begin{aligned} \sin (θ) = - \frac{15}{17} & = \frac{y}{r} \\ ↓ \\ \frac{y}{r} & = - \frac{15}{17} \\ This means: \\ y & = - 15 \\ r & = 17 \end{aligned}

$\begin{aligned} \sin (\theta) = - \dfrac{15}{17} & = \dfrac{y}{r} \\ & \downarrow \\ \dfrac{y}{r} & = - \dfrac{15}{17} \\ \text{This means:} & \\ y & = -15 \\ r & = 17 \end{aligned}$

STEP: Find the value of

x

$x$
[−1 point ⇒ 1 / 5 points left]

To find the ratio we want, we need the value of $x$ . We can find this using the theorem of Pythagoras for the reference triangle in the figure.

\begin{aligned} a^{2} + b^{2} & = c^{2} \\ x^{2} + (15)^{2} & = (17)^{2} \\ x^{2} & = 289 - 225 \\ x^{2} & = 64 \\ x & = \pm 8 \end{aligned}

$\begin{aligned} a^{2} + b^{2} & = c^{2} \\ x^{2} + (15)^{2} & = (17)^{2} \\ x^{2} & = 289 - 225 \\ x^{2} & = 64 \\ x & = \pm 8 \end{aligned}$

For the third side of the triangle, we must take the positive value (because it is a distance). However, the sign of $x$ depends on which quadrant the point is in. In this case, Point $P$ is in Quadrant III, where the $x$ -values must be negative. So the correct value for the $x$ -coordinate is $-\text{8}$ (we throw away the positive answer).

STEP: Use the definition of the tangent ratio on the Cartesian plane
[−1 point ⇒ 0 / 5 points left]

Based on the calculation above we can complete the information about the triangle:

Since we know that Point $P$ corresponds to angle $\alpha$ , we can evaluate $\tan (\alpha)$ by substituting in the correct values from the point.

\begin{aligned} \tan (θ) & = \frac{y}{x} \\ = \frac{15}{8} \end{aligned}

$\begin{aligned} \tan (\theta) & = \dfrac{y}{x} \\ & = \dfrac{15}{8} \end{aligned}$

TIP: Point

P

$P$ is in Quadrant III, where the CAST diagram tells us that the tangent ratio is always positive. Make sure your answer agrees with that!

The value of $\tan (\alpha)$ is $\dfrac{15}{8}$ .

Submit your answer as:

ID is: 3596 Seed is: 7953

Working with trigonometric ratios

The diagram below shows Point $P$ at $(3; 7)$ . Triangle $OPQ$ is also shown, with three sides labelled as $\text{3}$ , $\text{7}$ , and $\sqrt{58}$ . The angle $\omega$ reaches from the positive $x$ -axis to the line $OP$ , as labelled. Answer the two questions which follow below.

Determine the value of $\tan \omega$ .

INSTRUCTION: Your answer should be exact. If the answer includes a surd, you should type it using sqrt( ).

Answer: $\tan \omega =$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the definition for the tangent ratio on the Cartesian plane:

$\begin{array}{c} \tan \theta = \dfrac{y}{x} \end{array}$

STEP: Use the definition of the tangent ratio on the Cartesian plane
[−2 points ⇒ 0 / 2 points left]

On the Cartesian plane the tangent ratio is defined as $\dfrac{y}{x}$ . That means that we can read the answer from the diagram in the question. We need the values for $y$ and $x$ for Point $P$ .

$P$ is at $(3; 7)$ . So we get:

$\begin{array}{c} y = 7 \quad \text{and} \quad x = 3 \end{array}$

Putting those numbers into the definition for the tangent ratio, we get:

$\begin{alignedat}{1} \tan \omega & = \dfrac{y}{x} \\ & \downarrow \\ \tan \omega & = \dfrac{7}{3} \end{alignedat}$

NOTE: The size of angle $\omega$ puts Point $P$ in Quadrant I. The CAST diagram tells us that the tangent ratio is always positive in Quadrant I. And indeed the answer we got agrees with that: it is positive because both coordinates are positive.

The value of $\tan \omega$ is $\dfrac{7}{3}$ .

Submit your answer as:
Hence evaluate the following expression:

$\begin{array}{c} 2\tan \omega-\dfrac{4}{5} \end{array}$

Answer: $2\tan \omega-\dfrac{4}{5} =$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by substituting the answer from Question 1 into the expression. Then evaluate the expression.

STEP: Substitute the value of $\tan \omega$ into the expression
[−1 point ⇒ 1 / 2 points left]

Start by substituting the answer from Question 1 into the expression:

$\begin{aligned} & 2\tan \omega-\dfrac{4}{5} \\ & = 2\left(\dfrac{7}{3}\right)-\dfrac{4}{5} \end{aligned}$

STEP: Evaluate the expression
[−1 point ⇒ 0 / 2 points left]

Now evaluate the expression.

$\begin{aligned} & 2\left(\dfrac{7}{3}\right)-\dfrac{4}{5} \\ & = \dfrac{14}{3}-\dfrac{4}{5} \\ & = \dfrac{ \left(14 \cdot 5\right) + \left(-4 \cdot 3\right) }{ 3 \cdot 5 } \\ & = \frac{58}{15} \end{aligned}$

The value of $2\tan \omega-\dfrac{4}{5}$ is $\dfrac{58}{15}$ .

Submit your answer as:

ID is: 3596 Seed is: 8385

Working with trigonometric ratios

The diagram below shows Point $P$ at $(-3; 7)$ . Triangle $OPQ$ is also shown, with three sides labelled as $\text{3}$ , $\text{7}$ , and $\sqrt{58}$ . The angle $\theta$ reaches from the positive $x$ -axis to the line $OP$ , as labelled. Answer the two questions which follow below.

Determine the value of $\sin \theta$ .

INSTRUCTION: Your answer should be exact. If the answer includes a surd, you should type it using sqrt( ).

Answer: $\sin \theta =$
expression
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the definition for the sine ratio on the Cartesian plane:

$\begin{array}{c} \sin \theta = \dfrac{y}{r} \end{array}$

STEP: Use the definition of the sine ratio on the Cartesian plane
[−2 points ⇒ 0 / 2 points left]

On the Cartesian plane the sine ratio is defined as $\dfrac{y}{r}$ . That means that we can read the answer from the diagram in the question. We need the values for $y$ and $r$ for Point $P$ .

$P$ is at $(-3; 7)$ . From the diagram we can see that the radius is $r = \sqrt{58}$ . So we get:

$\begin{array}{c} y = 7 \quad \text{and} \quad r = \sqrt{58} \end{array}$

Putting those numbers into the definition for the sine ratio, we get:

$\begin{alignedat}{1} \sin \theta & = \dfrac{y}{r} \\ & \downarrow \\ \sin \theta & = \dfrac{7}{\sqrt{58}} \end{alignedat}$

NOTE: The size of angle $\theta$ puts Point $P$ in Quadrant II. The CAST diagram tells us that the sine ratio is always positive in Quadrant II. And indeed the answer we got agrees with that: it is positive because the $y$ -coordinate is positive (and the radius is always positive).

The value of $\sin \theta$ is $\dfrac{7}{\sqrt{58}}$ .

Submit your answer as:
Hence evaluate the following expression:

$\begin{array}{c} 2\sin ^2\theta-\dfrac{3}{2} \end{array}$

Answer: $2\sin ^2\theta-\dfrac{3}{2} =$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by substituting the answer from Question 1 into the expression. Then evaluate the expression.

STEP: Substitute the value of $\sin \theta$ into the expression
[−1 point ⇒ 1 / 2 points left]

Start by substituting the answer from Question 1 into the expression:

$\begin{aligned} & 2\sin ^2\theta-\dfrac{3}{2} \\ & = 2\left(\dfrac{7}{\sqrt{58}}\right)^2-\dfrac{3}{2} \end{aligned}$

STEP: Evaluate the expression
[−1 point ⇒ 0 / 2 points left]

Now evaluate the expression.

$\begin{aligned} & 2\left(\dfrac{7}{\sqrt{58}}\right)^2-\dfrac{3}{2} \\ & = 2\cdot \dfrac{49}{58}-\dfrac{3}{2} \\ & = \dfrac{49}{29}-\dfrac{3}{2} \\ & = \dfrac{ \left(49 \cdot 2\right) - \left(3 \cdot 29\right) }{ 29 \cdot 2 } \\ & = \frac{11}{58} \end{aligned}$

The value of $2\sin ^2\theta-\dfrac{3}{2}$ is $\dfrac{11}{58}$ .

Submit your answer as:

ID is: 3596 Seed is: 5280

Working with trigonometric ratios

The diagram below shows Point $P$ at $(3; -5)$ . Triangle $OPQ$ is also shown, with three sides labelled as $\text{3}$ , $\text{5}$ , and $\sqrt{34}$ . The angle $\omega$ reaches from the positive $x$ -axis to the line $OP$ , as labelled. Answer the two questions which follow below.

Determine the value of $\cos \omega$ .

INSTRUCTION: Your answer should be exact. If the answer includes a surd, you should type it using sqrt( ).

Answer: $\cos \omega =$
expression
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the definition for the cosine ratio on the Cartesian plane:

$\begin{array}{c} \cos \theta = \dfrac{x}{r} \end{array}$

STEP: Use the definition of the cosine ratio on the Cartesian plane
[−2 points ⇒ 0 / 2 points left]

On the Cartesian plane the cosine ratio is defined as $\dfrac{x}{r}$ . That means that we can read the answer from the diagram in the question. We need the values for $x$ and $r$ for Point $P$ .

$P$ is at $(3; -5)$ . From the diagram we can see that the radius is $r = \sqrt{34}$ . So we get:

$\begin{array}{c} x = 3 \quad \text{and} \quad r = \sqrt{34} \end{array}$

Putting those numbers into the definition for the cosine ratio, we get:

$\begin{alignedat}{1} \cos \omega & = \dfrac{x}{r} \\ & \downarrow \\ \cos \omega & = \dfrac{3}{\sqrt{34}} \end{alignedat}$

NOTE: The size of angle $\omega$ puts Point $P$ in Quadrant IV. The CAST diagram tells us that the cosine ratio is always positive in Quadrant IV. And indeed the answer we got agrees with that: it is positive because the $x$ -coordinate is positive (and the radius is always positive).

The value of $\cos \omega$ is $\dfrac{3}{\sqrt{34}}$ .

Submit your answer as:
Hence evaluate the following expression:

$\begin{array}{c} 4\cos ^2\omega-\dfrac{4}{3} \end{array}$

Answer: $4\cos ^2\omega-\dfrac{4}{3} =$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by substituting the answer from Question 1 into the expression. Then evaluate the expression.

STEP: Substitute the value of $\cos \omega$ into the expression
[−1 point ⇒ 1 / 2 points left]

Start by substituting the answer from Question 1 into the expression:

$\begin{aligned} & 4\cos ^2\omega-\dfrac{4}{3} \\ & = 4\left(\dfrac{3}{\sqrt{34}}\right)^2-\dfrac{4}{3} \end{aligned}$

STEP: Evaluate the expression
[−1 point ⇒ 0 / 2 points left]

Now evaluate the expression.

$\begin{aligned} & 4\left(\dfrac{3}{\sqrt{34}}\right)^2-\dfrac{4}{3} \\ & = 4\cdot \dfrac{9}{34}-\dfrac{4}{3} \\ & = \dfrac{18}{17}-\dfrac{4}{3} \\ & = \dfrac{ \left(18 \cdot 3\right) - \left(4 \cdot 17\right) }{ 17 \cdot 3 } \\ & = - \frac{14}{51} \end{aligned}$

The value of $4\cos ^2\omega-\dfrac{4}{3}$ is $- \dfrac{14}{51}$ .

Submit your answer as:

ID is: 3600 Seed is: 9011

Using the CAST diagram

The figure below shows the CAST diagram. Point $P$ is shown, and angle $\theta$ is the angle from the positive $x$ -axis to Point $P$ .

Is $\tan \theta$ positive or negative?

Answer: $\tan \theta$ is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Point $P$ is in Quadrant III. What does the CAST diagram tell you about the sign of the tangent ratio in that quadrant?

STEP: Read the answer from the CAST diagram
[−1 point ⇒ 0 / 1 points left]

We need to determine if the expression $\tan \theta$ is positive or negative. We can find the answer using the CAST diagram because the CAST diagram tells us when the trigonometric ratios are positive or negative.

We can see that the angle $\theta$ puts Point $P$ in Quadrant III. On the CAST diagram, what letter is in Quadrant III? It is T. The T means that the tangent ratio is positive in that quadrant and the other ratios are negative.

Based on the CAST diagram, the expression $\tan \theta$ must be positive.

Submit your answer as:
Now determine if $\cos \theta$ is positive or negative.

Answer: $\cos \theta$ is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the same approach as you used in Question 1.

STEP: Read the answer from the CAST diagram
[−1 point ⇒ 0 / 1 points left]

For this question, we use exactly the same information that we used in Question 1. Point $P$ and $\theta$ are in Quadrant III, where we find the letter T on the CAST diagram. This means that the tangent ratio is positive for any angle in that quadrant, while cosine and sine are negative.

NOTE:
What do the answers to these questions mean?

The angle shown in the diagram is $\theta = 234°$ . We can calculate the value of the expressions for these two questions (use a calculator):

$\begin{alignedat}{1} \text{From } & \text{Question 1}: \\ \tan 234° & = \text{1.37638...} \\[0.5cm] \text{From } & \text{Question 2}: \\ \cos 234° & = -\text{0.58778...} \end{alignedat}$

You can see that these values are positive and negative, just as we already found using the CAST diagram. That is what the answers to these question mean: they tell us the signs of the answers even if we cannot know the answers themselves!

The expression $\cos \theta$ must be negative.

Submit your answer as:

ID is: 3600 Seed is: 1083

Using the CAST diagram

The figure below shows the CAST diagram. Point $P$ is shown, and angle $\theta$ is the angle from the positive $x$ -axis to Point $P$ .

Is $\tan \theta$ positive or negative?

Answer: $\tan \theta$ is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Point $P$ is in Quadrant I. What does the CAST diagram tell you about the sign of the tangent ratio in that quadrant?

STEP: Read the answer from the CAST diagram
[−1 point ⇒ 0 / 1 points left]

We need to determine if the expression $\tan \theta$ is positive or negative. We can find the answer using the CAST diagram because the CAST diagram tells us when the trigonometric ratios are positive or negative.

We can see that the angle $\theta$ puts Point $P$ in Quadrant I. On the CAST diagram, what letter is in Quadrant I? It is A. The A means that all three of the trigonometric ratios are positive for angles in Quadrant I.

Based on the CAST diagram, the expression $\tan \theta$ must be positive.

Submit your answer as:
Now determine if $\cos \theta$ is positive or negative.

Answer: $\cos \theta$ is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the same approach as you used in Question 1.

STEP: Read the answer from the CAST diagram
[−1 point ⇒ 0 / 1 points left]

We already know the answer from Question 1: for angles in Quadrant I the trigonometric ratios are always positive. It does not even matter which ratio we are talking about! If the angle is between $0°$ and $90°$ (in Quadrant I) then the value of the ratio will be positive

NOTE:
What do the answers to these questions mean?

The angle shown in the diagram is $\theta = 44°$ . We can calculate the value of the expressions for these two questions (use a calculator):

$\begin{alignedat}{1} \text{From } & \text{Question 1}: \\ \tan 44° & = \text{0.96568...} \\[0.5cm] \text{From } & \text{Question 2}: \\ \cos 44° & = \text{0.71933...} \end{alignedat}$

You can see that these values are both positive, just as we already found using the CAST diagram. That is what the answers to these question mean: they tell us the signs of the answers even if we cannot know the answers themselves!

The expression $\cos \theta$ must be positive.

Submit your answer as:

ID is: 3600 Seed is: 8548

Using the CAST diagram

The figure below shows the CAST diagram. Point $P$ is shown, and angle $\theta$ is the angle from the positive $x$ -axis to Point $P$ .

Is $\tan \theta$ positive or negative?

Answer: $\tan \theta$ is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Point $P$ is in Quadrant IV. What does the CAST diagram tell you about the sign of the tangent ratio in that quadrant?

STEP: Read the answer from the CAST diagram
[−1 point ⇒ 0 / 1 points left]

We need to determine if the expression $\tan \theta$ is positive or negative. We can find the answer using the CAST diagram because the CAST diagram tells us when the trigonometric ratios are positive or negative.

We can see that the angle $\theta$ puts Point $P$ in Quadrant IV. On the CAST diagram, what letter is in Quadrant IV? It is C. The C means that the cosine ratio is positive in that quadrant and the other ratios are negative.

Based on the CAST diagram, the expression $\tan \theta$ must be negative.

Submit your answer as:
Now determine if $\cos \theta$ is positive or negative.

Answer: $\cos \theta$ is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the same approach as you used in Question 1.

STEP: Read the answer from the CAST diagram
[−1 point ⇒ 0 / 1 points left]

For this question, we use exactly the same information that we used in Question 1. Point $P$ and $\theta$ are in Quadrant IV, where we find the letter C on the CAST diagram. This means that the cosine ratio is positive for any angle in that quadrant, while sine and tangent are negative.

NOTE:
What do the answers to these questions mean?

The angle shown in the diagram is $\theta = 303°$ . We can calculate the value of the expressions for these two questions (use a calculator):

$\begin{alignedat}{1} \text{From } & \text{Question 1}: \\ \tan 303° & = -\text{1.53986...} \\[0.5cm] \text{From } & \text{Question 2}: \\ \cos 303° & = \text{0.54463...} \end{alignedat}$

You can see that these values are negative and positive, just as we already found using the CAST diagram. That is what the answers to these question mean: they tell us the signs of the answers even if we cannot know the answers themselves!

The expression $\cos \theta$ must be positive.

Submit your answer as:

$x$	$f(x)$
$0°$	$1$
$60°$	$\frac{1}{2}$
$120°$	$?$
$180°$	$-1$
$240°$	$-\frac{1}{2}$
$300°$	$\frac{1}{2}$
$360°$	$1$

$x$	$f(x)$
$0°$	$0$
$90°$	$?$
$180°$	$0$
$270°$	$\text{undefined}$
$360°$	$0$

Trigonometry

1. Trigonometry

Working with trigonometric equations: cosine

Working with trigonometric equations: cosine

Working with trigonometric equations: cosine

Trigonometry: using a calculator

Trigonometry: using a calculator

Trigonometry: using a calculator

Working with trigonometric equations: tangent

Working with trigonometric equations: tangent

Working with trigonometric equations: tangent

Trigonometry with non-right triangles

Trigonometry with non-right triangles

Trigonometry with non-right triangles

Working with trigonometric equations: sine

Working with trigonometric equations: sine

Working with trigonometric equations: sine

2. Special angles

Proportionality in the trigonometric ratios

Proportionality in the trigonometric ratios

Proportionality in the trigonometric ratios

Special angles and equations

Special angles and equations

Special angles and equations

The special angles

The special angles

The special angles

Special angle trigonometric ratios

Special angle trigonometric ratios

Special angle trigonometric ratios

The special angles

The special angles

The special angles

Special angles: what is the angle?

Special angles: what is the angle?

Special angles: what is the angle?

Identifying special angles

Identifying special angles

Identifying special angles

Memorise those special angle ratios!

Memorise those special angle ratios!

Memorise those special angle ratios!

Using the special angles to solve problems

Using the special angles to solve problems

Using the special angles to solve problems

Trigonometry: special angles

Trigonometry: special angles

Trigonometry: special angles

Special angles and triangles

Special angles and triangles

Special angles and triangles

3. Practical applications

4. Trigonometry on a unit circle

Trigonometry on the Cartesian plane

Trigonometry on the Cartesian plane

Trigonometry on the Cartesian plane

The trigonometric ratios on the Cartesian plane

The trigonometric ratios on the Cartesian plane

The trigonometric ratios on the Cartesian plane

Reading ratios from the Cartesian plane

Reading ratios from the Cartesian plane

Reading ratios from the Cartesian plane

Simplifying trigonometric expressions with reduction formulas

Simplifying trigonometric expressions with reduction formulas

Simplifying trigonometric expressions with reduction formulas

Special angles: 0°0°0° and 90°90°90°

Special angles: 0°0°0° and 90°90°90°

Special angles: 0°0°0° and 90°90°90°

Special angles: 180°180°180°, 270°270°270°, and 360°360°360°

Special angles: 180°180°180°, 270°270°270°, and 360°360°360°

Special angles: 180°180°180°, 270°270°270°, and 360°360°360°

Trigonometry gone wild!

Trigonometry gone wild!

Trigonometry gone wild!

Getting familiar with the CAST diagram

Getting familiar with the CAST diagram

Getting familiar with the CAST diagram

Transferring trigonometric ratios to the Cartesian plane

Transferring trigonometric ratios to the Cartesian plane

Transferring trigonometric ratios to the Cartesian plane

Special angles: $0°$ and $90°$

Special angles: $0°$ and $90°$

Special angles: $0°$ and $90°$

Special angles: $180°$ , $270°$ , and $360°$

Special angles: $180°$ , $270°$ , and $360°$

Special angles: $180°$ , $270°$ , and $360°$